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thermodynamics
|
Mass fraction of fuel on a fuel droplet surface
|
https://physics.stackexchange.com/questions/66766/mass-fraction-of-fuel-on-a-fuel-droplet-surface
|
<p>Concerning combustion of fuel droplets:</p>
<p>Why is the mass fraction of fuel on a fuel droplet surface slightly less than one?</p>
<p>It is known that the temperature is below the boiling point at the fuel droplet surface, so there is no evaporation at the surface and therefore my intuition tell me that the mass fraction of the fuel at the surface should be unity, but it isn't. So why is the mass fraction of fuel less than one, and what happened to that fraction of fuel mass?</p>
|
<p>At the liquid surface, the vapor phase exists as a mixture of fuel vapor ($m_f$ - which is a function of surface temperature, $T_s$) and air ($m_a$). As a result the mass-fraction of fuel vapor ($Y_{f,s}$) at the fuel surface will be less than unity. Depending on the volatility of the fuel and the ambient conditions, the value of $Y_{f,s}$ will vary from a value slightly below unity to zero (especially when the ambient temperature too low compared to the normal boiling point of the fuel).</p>
| 200
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thermodynamics
|
Is there a source that provides data for the temperature coefficient of resistivity at different temperatures?
|
https://physics.stackexchange.com/questions/67977/is-there-a-source-that-provides-data-for-the-temperature-coefficient-of-resistiv
|
<p>I'm looking for a source where I can find the temperature coefficient of resistivity at different temperatures for pure metals. Tables are everywhere for 20$\,^{\circ}$C, but I'm having difficulties finding data for other temperatures.</p>
<p>One resource I looked at (<a href="http://www.physicsforums.com/showthread.php?t=546194" rel="nofollow">http://www.physicsforums.com/showthread.php?t=546194</a>) provides a formula:</p>
<p>$\alpha_T = \left(\frac{1}{\frac{1}{\alpha_0} + T}\right)$</p>
<p>I don't believe that's applicable to all materials, but is there a group of materials where that expression is true?</p>
<hr>
<p><strong>[UPDATE]</strong></p>
<p>@Trimok led me to the Bloch–Grüneisen formula</p>
<p>$\rho(T)=\rho(0)+A\left(\frac{T}{\Theta_R}\right)^n\int_0^{\frac{\Theta_R}{T}}\frac{x^n}{(e^x-1)(1-e^{-x})}dx$</p>
<p>I can work with that, but there's a lot of parameters that I would need before solving it. Is there a source for $\rho\left(0\right)$, $A$, $\Theta_R$ (the Debye temperature), for different metals? Also which of the n values need to be considered?</p>
| 201
|
|
thermodynamics
|
Monotherm cycle, negative $Q_{net}$
|
https://physics.stackexchange.com/questions/69462/monotherm-cycle-negative-q-net
|
<p>In the demonstration of the Clausius theorem in classical thermodynamics my book uses the fact that $Q_{net}\le0$ is negative in a cyclic monotherm (meaning with one heat reservoir) transformation. This fact, however, left me a bit uneasy. Taking, as usual, a Carnot cycle as an example, and calling the adiabatic and the isotherm expansions done, couldn't I just do an adiabatic compression to bring it back to the initial state? The second law wouldn't be violated: the heat absorbed wouldn't <em>entirely</em> be converted into work. The expression of the first principle, though, is indeed a bit strange: since $\Delta U$ is $0$ in any cycle, then $Q = W$, and my reasoning breaks down. I can't really see what's going on here, would it be theoretically possible a process similar to the one I described? If not, why so?</p>
|
<p>Let's assume that the thermodynamic state of the system can be represented in two dimensions (like on a pressure-volume diagram for an ideal gas). Let's say further that the cycle starts at point $a$, undergoes an isothermal expansion to point $b$, then undergoes an adiabatic expansion to point $c$.</p>
<p>As far as I understand, you now want to take the system from point $c$ to point $a$ by way of an adiabatic compression. This is not possible because there is only one adiabatic curve passing through point $c$; it is precisely the curve along which the system underwent the adiabatic expansion in the first place. There is no adiabatic curve connecting $a$ and $c$.</p>
| 202
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thermodynamics
|
Can I run a pulse tube cooler in reverse to generate pressure oscillations?
|
https://physics.stackexchange.com/questions/71000/can-i-run-a-pulse-tube-cooler-in-reverse-to-generate-pressure-oscillations
|
<p>Since a <a href="http://en.wikipedia.org/wiki/Pulse_tube_refrigerator" rel="nofollow">pulse tube cooler</a> is basically a Stirling engine without moving parts running "in reverse", does that imply that when I keep the cold end at ambient temperature and apply heat to the hot end, I get pressure oscillations? Or do I need the pistons and crankshaft of a standard Stirling engine for that?</p>
<p>I think the question can be boiled down to how the working fluid gets back to the hot end after cooling, and the reason I don't seem to understand this might be that I have not yet fully understood how the buffer volume and orifice work in the pulse tube cooler.</p>
| 203
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thermodynamics
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Energy provided to piston in compression and exhaust stroke?
|
https://physics.stackexchange.com/questions/72574/energy-provided-to-piston-in-compression-and-exhaust-stroke
|
<p>In the compression stroke of a petrol engine, the mixture is compressed by the upward movement of the piston. From where does the piston get energy to compress the mixture?</p>
<p>Similarly, in the exhaust stroke, the piston again moves upward to expel the gases. Where does the piston get energy from?</p>
|
<p>Using directions like <i>upwards</i> in this context is meaningless because you haven't defined how the piston is oriented.</p>
<p>During the compression stroke of a typical 4-stroke gasoline engine, the piston takes energy from the crankshaft. The reason the whole scheme still works is that you get a lot more energy back after the cumbustion during the power stroke. Add up the total energy moved between the piston and the crankshaft over a whole cycle (two rotations of the crankshaft), and you will find the total being positive from piston to crankshaft.</p>
<p>The exhaust stroke takes relatively little energy since in theory the exhaust valve is open during this time and little pressure is required to expell the exhaust gasses.</p>
<p>You missed the remaining stroke, which is intake. That also takes a little energy since the pressure in the piston is negative in a traditional engine during that time. The piston is sucking the air or air-fuel mixture into the cylinder during that time.</p>
<p>In summary, the intake, compression, and exhaust strokes all take some energy from the rotation of the crankshaft, but this is more than made up by the energy imparted onto the crankshaft during the power stroke. This is also one reason these engines need to be started mechanically. It takes at least two energy-robbing strokes before you get the first positive return. This initial energy has to be supplied externally by rotating the crankshaft, such as is done by the starter motor in your car or by your arm when you pull the ripcord of your lawnmower.</p>
<p>At best, a piston is supplying energy only 1/4 of the time. This is why 4 cylinders is a common number for multi-cylinder engines. One piston is on its power stroke. Some of this energy is used to power the other three cylinders, with the rest being the output of the engine. More cylinders makes the overall operation smoother. More of a flywheel is needed with less cylinders, especially when you get to less than 4.</p>
| 204
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thermodynamics
|
Can water exist as liquid in a vacuum chamber at room temperature?
|
https://physics.stackexchange.com/questions/74490/can-water-exist-as-liquid-in-a-vacuum-chamber-at-room-temperature
|
<p>I know boiling point at vacuum is far below the room temperature. But I still wondering if the humidity reaches the saturated humidity (100% relative humidity), will the vapor condense to liquid, although it is probably going to evaporate again soon?</p>
|
<p>While dmckee points out your question is too vague, I'll take a shot and guess you're asking from a pop-sci standpoint. Since you mention vacuum chamber, I'm going to assume it's actively maintained. For example put a wet shirt in a vacuum chamber and pump it down.</p>
<p>The water will boil off but it has to be removed from the system because you are trying to maintain a vacuum. Thus the pump will condense the water (along with other gas molecules) outside of the chamber. By the time you get down to a good low vacuum pressure (< 1 mTorr) there will be no water or very few molecules left. Release the vacuum and your shirt will be dry.</p>
| 205
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thermodynamics
|
Water coolso object by heat absorption
|
https://physics.stackexchange.com/questions/76367/water-coolso-object-by-heat-absorption
|
<p>Why do water of water based liquids like sweat cool objects? The most clear example I have is the sweat on the skin. I learned that it absorbed heat because it evaporates, but this is something strange. I also read that by putting a little water on a can under the sun will cool the inside of the can, for the same reason as sweat. Why is this?
Why does a liquid take away the heat, while it isn't even at the boiling point? The body is only 37 degrees Celcius, why does sweat "want" to evaporate already? I do know that liquids take energy to change its phase without adding to the temperature, but this is only at boiling point.</p>
|
<p>Well you almost got it but then talked yourself out of it. Yes it does take energy (latent heat) to convert a liquid (water) to a vapor . That is the energy to overcome thing like Van der Waals forces that hold close packed molecules together and allow them to float free of such constraints.</p>
<p>But it is NOT true, that this only happens at the Boiling point. It happens at any Temperature. It takes approximately 593 calories per gram of water (at 100 deg C) to convert it into steam at 100 deg C.
At a lower Temperature, the water molecules will have less average kinetic energy, so it will take more energy to convert it to vapor at the lower Temperature, so the latent heat of evaporation will be somewhat higher than 593 calories per gram.
As a rough figure, you can assume one more calorie per gram for each degree lower Temperature. So for ice water at near zero deg C it will take about 693 Calories per gram.</p>
<p>If the water is on your skin, that latent heat will be extracted from your warmer skin, which is why it makes you feel cooler.</p>
<p>The Temperature of a mass of material, reflects the average kinetic energy per molecule of the material. But individual molecules have diffent energies which they exchange during their collisions. Thes energies generally are distributed with some statistical distribution, often a Maxwell-Boltzmann distribution, which has a long high energy tail to it. So there always are molecules with sufficient energy to escape the surface of the liquid, and fly free in the gaseous phase. This lowers the average KE of the remainder , so it lowers the Temperature of the liquid. But more high energy tail molecules will escape, anyhow, so evaporation can take place even well below the boiling point. Of course if the atmosphere above the liquid should start to have more molecues of the liquid, (higher humidity for example), then some of those can collide with the liquid surface and be re-absorbed. Evaporation will stop, when the number of higher energy molecules escaping, equals the number returning from the atmosphere above the liquid.</p>
<p>Hurricanes, which are powered by heat from warmer waters, suck a lot of latent heat of evaporation out of the ocean surface, so the hurricane leaves a cold surface water track behind it.</p>
| 206
|
thermodynamics
|
Thermodynamics in a closed room - Why does the stale smell depend on Temperature?
|
https://physics.stackexchange.com/questions/78620/thermodynamics-in-a-closed-room-why-does-the-stale-smell-depend-on-temperature
|
<p>Why does the stale smell disappears in a crowded room if you cool it via air-condition ? What's the physics ?</p>
|
<p>It doesn't have anything to do with thermodynamics – the stale smell is due to certain molecules or mould spores floating in the air. Putting fresh air into the room, and removing the old air, displaces these floating molecules and removes the smell.</p>
| 207
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thermodynamics
|
How can you tell between gas, liquid and crystal in a microscopic way?
|
https://physics.stackexchange.com/questions/82388/how-can-you-tell-between-gas-liquid-and-crystal-in-a-microscopic-way
|
<p>I know that molecules in ideal gas can move freely, and molecules in crystal are bonded to some specific location. But can I describe this in a more quantitative way? Do gas molecules have more degrees of freedom?</p>
|
<p>Yes, by examining the statistical distribution of distances between molecules and angles separating two nearby about a third molecule. In general, correlations of 2nd and higher order of the positions of molecules relative to each other.</p>
<p>For a gas, there are few molecules close together, but some due to molecules colliding and almost colliding. At far distances, it'll be more or less uniform. There'd be nothing of interest in angular correlations.</p>
<p>For a liquid, there'd be none closer than about the size of a molecule, but at that distance many. There'd be mushy peaks in the distribution of distances, and just uniform mush beyond a few molecule-sizes away. There'd be strong angular correlations as nearby molecules try to pack tightly, with fleeting gatherings of several molecules in an approximate crystal, but always jiggling making the angular distribution mushy.</p>
<p>For a crystalline solid, every molecule near or far is at a precise distance, and at precise angles with respect to any reference directions. The distance and angular distributions would look like bunches of Dirac functions, slightly smoothed out due to thermal motion, phonons, impurites and so on.</p>
<p><a href="http://rkt.chem.ox.ac.uk/lectures/liqsolns/liquids.html" rel="nofollow">Radial distributions explained by professors at Oxford, with plots</a></p>
<p><a href="http://matdl.org/matdlwiki/index.php/softmatter%3aRadial_Distribution_Function" rel="nofollow">Comparison of radial distribution functions, with plots</a></p>
| 208
|
thermodynamics
|
Experimental heat equation for a bottle of water
|
https://physics.stackexchange.com/questions/83956/experimental-heat-equation-for-a-bottle-of-water
|
<p>I'm not sure if this question belongs here, because I'm asking a question about experimental data rather than theoretic formula.</p>
<p>Anyway, I'm trying to figure out the time-temperature relation of the (hot) water in an ordinary sports bottle. Assume the bottle is full with 500mL of water in it. Do you guys have any data for that? The equation doesn't have to be accurate, an error up to 20% can be accepted.</p>
| 209
|
|
thermodynamics
|
Definition and Measurment of Energy and Heat in Thermodynamics
|
https://physics.stackexchange.com/questions/87222/definition-and-measurment-of-energy-and-heat-in-thermodynamics
|
<p>The concept of work is clear from mechanics, where $dW=F\cdot dx$, leading to $dW=PdV$. Also different forms of energy are defined in mechanics, e.g., $E=\frac{1}{2}m v^2$, etc. How are the energy and heat defined and measured in thermodynamics?</p>
<p>Because once we define them independently, we can state the first law of thermodynamics, $d E=\delta Q+\delta W$.</p>
| 210
|
|
thermodynamics
|
What are the experimental observations behind the first principle of thermodynamics?
|
https://physics.stackexchange.com/questions/87080/what-are-the-experimental-observations-behind-the-first-principle-of-thermodynam
|
<p>As far as I understand it, the first principle of thermodynamics is a mere definition of the quantity “Heat”: $$\text d Q: = \text d L + \text d U.$$
This is somewhat the point of view taken in Fermi's introductory book "Thermodynamics": </p>
<blockquote>
<p>[...] $$\Delta U + L=0$$If the system is not thermally isolated, the first member of [eqn.] will be generally not equal to zero [...]
Substitute the [eqn.] with the more general: $$\Delta U + L = Q.$$
[...] Now we will call $Q$, by definition, the quantity of heat received by the system during the transformation.</p>
</blockquote>
<p>(if you want to read the full text you might want to google “Fermi Thermodynamics"... pag. 17).</p>
<p>I think that this point is logically sound and I have a quite good understanding of some of the above structure starting from here (e.g. the second principle). On the other hand I feel as I'm missing something. </p>
<p>To give an example, from mechanics, this is how I understand Newton's equation:</p>
<blockquote>
<p>It is a matter of fact that the positions and the velocities of a
mechanical system fully determine the accelerations of the system. Hence, the dynamic of each system follows second order differential equations: $$\ddot x = F(x,\dot x, t).$$</p>
</blockquote>
<p>An other example might be the second law of thermodynamics, that (in Clausius' form) is simply the statement of the fact that heat doesn't flow spontaneously from a cold body to an hotter one.</p>
<p>Since I find strange that something that is called a “principle” is a mere definition (after all, there's no assumption involved in making a definition), I ask: what are the experimental facts behind the first principle of thermodynamics?</p>
<p>Note: I understand that this is really about my personal understanding, however I think that this question can be useful to others. Furthermore, if something isn't clear and if I can improve my question, let me know.</p>
|
<p>Thermodynamics is a phenomenological description of macroscopic systems, and it's laws are based on empirical observations.</p>
<p>The first law, first states that a <strong>state function</strong> called <em>internal energy</em> $U$ exists for macroscopic systems (an experimental fact), that can be thought of as the analog of potential energy in mechanics, for macroscopic systems; then <em>defines</em> heat intake of the system:</p>
<p>1) For an adiabatically isolated macroscopic system (i.e., when the only sources of energy are mechanical), the amount of work required to change the state of the system only depends on the initial and final states. (an observational fact)</p>
<p>2) When the adiabatic constraint is removed the amount of work is no longer equal to the change in the internal energy, and their difference is <em>defined</em> as the heat intake of the system: (definition of heat)
$$\delta Q=dU-\delta W$$
Here, $\delta Q$ and $\delta W$ are not separately functions of state, but their sum (internal energy) is. Note that $\delta W$ is the work done <em>on</em> the system.</p>
| 211
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thermodynamics
|
Maxwell relations, confused on how the solution got to this answer
|
https://physics.stackexchange.com/questions/88605/maxwell-relations-confused-on-how-the-solution-got-to-this-answer
|
<p>I have a problem where, basically, in part (a) I correctly found a fundamental equation $$TdS = dE - 2\sigma l dx$$</p>
<p>Then the problem goes on to say that the only parameters of interest are $x$ and $T$, so I would assume we can write the relation $$dS =
\frac{\partial S}{\partial x} dx + \frac{\partial S}{\partial T}dT$$
by taking the total derivative. From this information, the solution goes on to say <strong>"We can read off the Maxwell relation</strong> $$\left(\frac{\partial S}{\partial x}\right)_T = \left(\frac{\partial(-2\sigma l)}{\partial T}\right)_x = -2l \frac{d \sigma}{dT} \qquad \sigma = \sigma_0 - \alpha T$$</p>
<p>Where $\sigma$ is given in the problem statement. I'm just not seeing where they find this maxwell relation....been stuck for quite some time. Any help would be appreciated, thanks. </p>
|
<p>I'll elaborate on my comment. Maxwell's relations are just observations that second partial derivatives of a function commute. So for some potential $\Phi(x,y)$, if you differentiate, $d\Phi = (\partial_x \Phi) dx + (\partial_y \Phi) dy$, then because second partial derivatives commute, i.e., $\partial_y \partial_x \Phi = \partial_x \partial_y \Phi$, we obtain an identity on the coefficients of $dx$ and $dy$ in $d\Phi$.</p>
<p>The Maxwell relation that the solution gives you is reminiscent of a potential $A$ that has differential</p>
<p>$dA = -S dT + 2 \sigma l dx$,</p>
<p>because if $\partial_T A = -S$ and $\partial_x A = 2\sigma l$, then $\partial_T \partial_x A = \partial_x A \partial_T A$ reproduces your identity.</p>
<p>Now, I am going about the problem from the wrong end, because I am given the solution and asked to find how to derive it. However, my suspicion is that you are supposed to recognize from the start that $dx$ looks like a change in volume term ($dV$) and $-2 \sigma l$ a measure of pressure ($P$). Once you have done this, you have many Maxwell relations at your disposal. The mathematics don't care about the exact form of $P dV$, in other words.</p>
| 212
|
thermodynamics
|
Thermodynamics and wall insulation
|
https://physics.stackexchange.com/questions/87662/thermodynamics-and-wall-insulation
|
<p>Australia's history has been littered with the problem of how to manage safety in bushfires. The local fire authority recommend consideration be given to preparing a bushfire management plan, deciding whether to stay and defend, or leave early.</p>
<p>A back- up plan is also suggested, which may include the use of a personal fire bunker.</p>
<p>Temperatures can be up to $935^\circ C$ in a bushfire, and above ground bunkers are marketed as being safe for one hour.</p>
<p>Fire loads can peak at $100kWm^{-2}$.</p>
<p>My question relates to the effectiveness of fire bunkers to provide safe shelter in a fire. Key aspects of this are the ability of the bunker to achieve the following criteria: 1. Non combustible construction 2. Limit heat flow through walls to $2.5kWm^{-2}$ 3. Limit max surface temperature on the inside walls to $75^\circ C$ 4. Limit air temperature on the inside walls to $40^\circ C$; 5. Fire rated door, with non-combustible seal;</p>
<p>Research into this topic has yielded the following: 1. English fire expert experienced in fire door testing and fire gasket seals says it can't be done; expect the people in the bunker to be microwaved; 2. USA makes tornado/hurricane safe refuges, nominally fire resistant, but Californian authority advice is: not to be used in a wildfire.</p>
<p>My question relates to the meeting of the nominated Aus standards. Some finite element studies I have done into items 1,2,3 says all three can be achieved.</p>
<p>Items 4 and 5 are problematic.</p>
<p>Item 4 is the internal air temp. Even with a heat flow reduction from $100kWm^{-2}$ to $2.5kWm^{-2}$, as nominated in the Aus guidelines, if there is a fire refuge in lockdown, with limited air mass, no ventilation, prelim studies into this show the temp will rise. Lots.</p>
<p>This is based on: $Q=mC_{p}\Delta T$</p>
<p>It is a one liner, but says the temp goes up. A lot. Not a lot of mass in the air. Lots of exposed area in the walls and roof.</p>
<p>With better insulation, we can get this heat input down to $0.5kWm^{-2}$. This is lower than sunlight.</p>
<p>But the Time to heat up above $40^\circ C$ from a base of $35^\circ C$ is still very low, based on this formula, much lower than 60 mins target.</p>
<p>As a structural engineer with many years in structures, who is doing this as a bit of a hobby, I am heading back to my Uni notes to help. No doubt I am missing something; or are the Brits/US authorities on the right track?</p>
<p>To complicate the issue, there are people in the bunker, all huffing and puffing as the temp rises. That complicates the physics a bit more.</p>
<p>Item 5 Heat seal; smoke seal; air seal. Again a bit problematic as fire doors distort and let air in. Break the seal. Also internal temperature rise also heats up the internal air pressure. Which would also break the seal.</p>
<hr>
<p>Trust this gives a bit of background into what is an interesting problem. Would appreciate people's feedback on possible ways to better evaluate the internal room temperature/pressure rise in the bunker.</p>
|
<p>I don't know if there is a reasonable technical solution to your problem, but I suspect that, from the point of view of physics, item 4 is manageable in principle: the inside of the bunker does not have to be empty. For example, you can keep a couple of tons of water in the bunker:-) You can also consider boiling latent heat for some liquid, but this can add some problems with pressure.</p>
| 213
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thermodynamics
|
Please describe how a vacuum flask/thermos works
|
https://physics.stackexchange.com/questions/89411/please-describe-how-a-vacuum-flask-thermos-works
|
<ol>
<li><p>Please describe how a <a href="http://en.wikipedia.org/wiki/Vacuum_flask" rel="nofollow">vacuum flask/thermos</a> works. </p></li>
<li><p>How does the vacuum prevent convection/conduction/radiation? </p></li>
<li><p>How does the lid with the curvy lines prevent either of the aforementioned heat transfers? </p></li>
<li><p>If there are any other parts that may prevent heat transfer, please also list them.</p></li>
</ol>
|
<p>A thermos flask holds the liquid in a pocket in the middle of the flask, that pocket is surrounded by a smaller glass envelope that surrounds the inner pocket. This glass envelope is hollow- it is a vacuum, this means there is no air in it, so it is a fantastic insulator. It keeps hot things hot and cold things cold.</p>
| 214
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thermodynamics
|
What is the difference between heat and work
|
https://physics.stackexchange.com/questions/89902/what-is-the-difference-between-heat-and-work
|
<p>If 700 J of work is applied onto a gas, and therefore the gas contracted 0.005 m^3 (from 0.01m^3), and the temperature stayed at a constant rate of 300 kelvin, then how come when pressure is added to the system (Not by adding more molecules, but instead by increasing the kinetic energy of the molecules by adding heat) until the system becomes adiabatic (The system has as much total energy as it did from the start), we don't say that the heat applied equals to 700J?</p>
<p>It would make perfect sense if 700J were added to a system, then in order for that system to return to the same state, 700J would be taken away.</p>
| 215
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thermodynamics
|
Heat balance: how to obtain $T(t)$ with energy in the form $u(T(t))$
|
https://physics.stackexchange.com/questions/90683/heat-balance-how-to-obtain-tt-with-energy-in-the-form-utt
|
<p>I have the heat balance equation for a cooling case in the form:</p>
<p>$$\frac{dU(T(t))}{dt}=-J$$</p>
<p>with</p>
<ul>
<li>$U$: energy (J)</li>
<li>$T$: temperature (K)</li>
<li>$t$: time (s)</li>
<li>$J$: heat flux leaving the system (W)</li>
</ul>
<p>I have $U$ from experimental data; $U$ is a function of sample temperature, $T$, measured in the experiment, both are functions of the experiment time $t$.</p>
<p>Now I want to solve the differential equation for $T(t)$; but the problem is that $U(T)$ is not definite because the sample releases crystallization heat and causes $U(T)$ to increase after initial decrease. </p>
<p>I was thinking of using only $U(t)$ because this function would be definite. But I need $T(t)$ at the end. </p>
<p>Do I need to go in 3D or are there other suggestions?</p>
|
<p>Your question cannot be answered in its present form. The RHS is simply the heat flux out of the system, and that will be different depending on what your system is surrounded by and how those surroundings <em>respond</em> to having the temperature of their boundary raised. </p>
<p>A rough solution is afforded by something like Newton's law of cooling: this describes the $J$ as being proportional to the temperature difference between the body and the "ambient" (i.e. many diffusion lengths into the surrounding material). So, if your heat content is linear over your temperature range of interest, so $U = \sigma T$, where $\sigma$ is your system's <em>heat capacity</em>, and if $k$ is the constant in Newton's law of cooling then:</p>
<p>$$\sigma \,{\rm d}_t \Delta U = - k\, \Delta U$$</p>
<p>where $\Delta U$ is the temperature difference between your body and the "ambient", whence $\Delta U(t) = \Delta U(0) \exp(-k\,t/\,\sigma)$. You would put your experimental data in for $U(T)$ instead of my simple $\sigma\,T$ into the above differential equation and solve accordingly. $k$ is highly dependent on what surrounds your body. Some insulating materials / boundary conditions will keep $k$ very small, maybe even effectively nought depending on what timescales / temperature scales you want your simulation to be good for. Other materials are excellent heat conductors and will give a very high $k$. So you see you need to know a great deal more about the surroundings of your body to make headway with your analysis.</p>
<p>Another level of sophistication is to think of the body and surroundings not as "lumped" elements each with a uniform temperature throughout each of them and account for diffustion. Then your equation becomes a partial DE for a scalar temperature field:</p>
<p>$$\nabla\cdot \vec{J}(\vec{r},t) = -\partial_t U(\vec{r},t)$$</p>
<p>You get this equation by replacing the LHS of your equation with the volume integral of $U(\vec{r},t)$ where $U(\vec{r},t)$ is a scalar field describing the heat content per unit volume. Then the time rate of change of this quantity is the surface integral $\oint_S \vec{J} \cdot \hat{\vec{n}} \,{\rm d}S$ of the heat flux in through the surface at the boundary of the volume. The convert the surface integral to a volume integral by Gauss's divergence threorem. I.e. here is where you would put in your experimental data $U(T(\vec{r},t))$. The "localised" Newton's law of cooling is then $ \vec{J} = -\kappa \nabla T(\vec{r},t)$ where $\kappa$ is the materials heat <em>conductivity</em> and, if you approximate $U(\vec{r},t) = \rho\,\sigma\,T(\vec{r},t)$ where now $\sigma$ is the mass-specific heat capacity and $\rho$ the density, we get the diffusion equation for homogeneous mediums:</p>
<p>$$\kappa \nabla^2 T - \sigma\,\rho\, T = 0$$</p>
<p>which is the diffusion equation. You then solve this equation for two mediums: your object, embedded in an infinite other representing the surroundings. The boundary conditions are that $T$ and the normal to surface component of $\kappa \nabla T$ must be continuous across the boundary (the latter is just conservation of heat flux across the boundary). </p>
| 216
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thermodynamics
|
Steam bath physics
|
https://physics.stackexchange.com/questions/90797/steam-bath-physics
|
<p>In the steam bath at the health club, why is the "steam" thicker first thing in the morning before it has been used all day and the walls are "hotter"?</p>
|
<p>When you say it's "thicker" in the morning, do you mean it seems more foggy/cloudy when the bath is first turned on? I suspect someone more well-versed in steam-bath design can refine this, but here is a guess:</p>
<p>When the steam bath is first turned on in the morning, steam is vented into a room whose air and walls are initially at ambient temperature. As steam vents into the room for the first few minutes, nucleation occurs in the air and condensation occurs against the walls, warming both. The thermal mass of the air in a steam bath room is considerably less than the thermal mass of the walls, so the air will probably approach thermal equilibrium on the timescale of 10 minutes or so, whereas the walls will probably take a couple hours to approach their steady-state temperature. So, in the morning, you have a room full of hot steam with cool walls, and later in the day, you have a room full of hot steam with hot walls.</p>
<p>In the morning, the steam-laden room air supplies heat to the walls via two routes, condensation and convective cooling. Condensation against the walls removes water from the air, and the thermal energy of condensation mostly goes into the walls. Convective cooling, however, does not reduce the water content of the air, but it reduces its temperature. Water is then forced to nucleate slowly (in order to equalize its chemical potential between phases), forming the fog present in the mornings.</p>
<p>In the afternoon, the walls will have more closely approached thermal equilibrium, and so the cooling of the air due to convective heat transfer will be reduced. As a result, there will be less nucleation, and the room air will not have as many microscopic water droplets floating around, making it seem "thinner".</p>
| 217
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thermodynamics
|
How does this Stirling engine work?
|
https://physics.stackexchange.com/questions/92642/how-does-this-stirling-engine-work
|
<p>I'm looking at <a href="http://books.google.ca/books?id=RN4_jLbVO3YC&pg=PA50&dq=Stirling+auto+engine+a+lot+of+progress,+but+Spirit&redir_esc=y#v=onepage&q&f=true" rel="nofollow">this</a> Stirling engine, which I believe is a variation of an alpha type. It was developed by Nasa back in the 1980's. The thing is, I can't figure out how the hydrogen cycles through the pistons. Could someone help me? This is another <a href="http://www.ohio.edu/mechanical/stirling/engines/engines.html" rel="nofollow">link</a> that might help you understand what it looks like if you scroll down about 3 quarters. There are several pictures on this page. I've tried making sense of it but I don't understand how the heated hydrogen cycles.</p>
| 218
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thermodynamics
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Thermo-couple Eelectroomotive force
|
https://physics.stackexchange.com/questions/92045/thermo-couple-eelectroomotive-force
|
<p>I am studying thermocouples. In a text book, the author said that, the electromotive force can be written as</p>
<p>$$ E= \alpha \theta + \beta \theta^2 \tag{1}$$</p>
<p>where $\alpha$ and $\beta$ are constants and the thermocouple is connected at $0 \;\mathrm{°C}$ on one end and at $100 \;\mathrm{°C}$ at the other end. </p>
<p>My problem is that I didn't find a good logic for the equation (1). Where does it arise from? </p>
|
<p>The voltage generated by a thermocouple arises from the <a href="http://en.wikipedia.org/wiki/Seebeck_effect#Seebeck_effect" rel="nofollow">Seebeck effect</a>. At the risk of over-simplifying, the electrons in a metal can be thought of as a gas of nearly free electrons, and when you heat a gas its pressure changes. So if you have a metal wire and there is a temperature difference between the ends electrons will tend to move from the hot end to the cold end. This sets up a potential difference, and this potential difference is what is measured in a thermocouple.</p>
<p>If the electrons behaved as an ideal gas the voltage would be simply proportional to the temperature difference, and indeed over small temperature ranges the voltage generated is approximately proportional to the temperature difference giving you:</p>
<p>$$ E \approx \alpha \theta $$</p>
<p>However it shouldn't surprise you to find that the electrons in a metal are only approximately free, and the potential difference is actually some complidated function of the temperature difference:</p>
<p>$$ E = f(\theta) $$</p>
<p>where $f(\theta)$ is the complicated function. Any function can be written as a <a href="http://en.wikipedia.org/wiki/Taylor_series" rel="nofollow">Taylor series</a>:</p>
<p>$$ f(\theta) = f(0) + \frac{f'(0)}{1!} \theta + \frac{f''(0)}{2!}\theta^2 + \frac{f'''(0)}{3!} \theta^3 + ...$$</p>
<p>This is an infinite series, but it's often the case that $\theta$ is small and so $\theta^2$ is even smaller and can be neglected. That gives us the linear approximation:</p>
<p>$$ f(\theta) \approx f(0) + \frac{f'(0)}{1!} \theta $$</p>
<p>Over a larger range of $\theta$ the value of $\theta^2$ becomes significant but $\theta^3$ is still small enough to be ignored and we get the quadratic approximation:</p>
<p>$$ f(\theta) \approx f(0) + \frac{f'(0)}{1!} \theta + \frac{f''(0)}{2!}\theta^2 $$</p>
<p>The quadratic approximation is being used in your formula for the voltage. The voltage is zero when the temperature is zero, so the first term $f(0)$ is zero and that leaves:</p>
<p>$$ E \approx \alpha \theta + \beta \theta^2 $$</p>
<p>where the constants $\alpha$ and $\beta$ are equal to $f'(0)$ and $f''(0)/2$ respectively. In practice the constants are empirical i.e. we would measure the voltage over a large range of temperatures and fit a quadratic equation to the experimental data.</p>
| 219
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thermodynamics
|
Would a cone shaped Stirling engine work?
|
https://physics.stackexchange.com/questions/92273/would-a-cone-shaped-stirling-engine-work
|
<p>A Stirling engine evidently functions by heating and cooling air, thus making the piston move up and down. What if the heated side of the cylinder were shaped as a sort of cone with a gentle slope, cut off before a point forms (I guess it would look like a trapezoid from the direct side), so that the surface that must be heated is reduced in size, lets say by half? Would that still permit the engine to run as well as it did with a large top surface?</p>
| 220
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thermodynamics
|
Does heat increase the volume of a gas, and in turn its pressure?
|
https://physics.stackexchange.com/questions/93072/does-heat-increase-the-volume-of-a-gas-and-in-turn-its-pressure
|
<p>Lets say you have 1 liter of hydrogen in a sealed container, at 100 psi. If 50 cm^2 of the containers surface area is heated to a 1000 degree Celsius, will the psi increase over time? What would be the formula to figure this out in terms of the amount of pressure added and the time frame it takes? No need to do the math for me, I'm capable of that, but I do need help with understanding which formula to use. Disregard the heat loss of the container for now.</p>
|
<p>The system will eventually reach thermal equilibrium so heat received by the gas is the same as heat lost by the container. So: $m c (T_f-T_i) + m c(T_f-T_i) = 0$. Hence you need to know the specific heat capacity of the materials, and the mass, as well as the initial temperatures. </p>
<p>The formula relating pressure, volume, amount of moles and temperature is: $PV = nRT$, where R is the gas constant (which depends on what units you use). From this you derive $P_1/T_1 = P_2/T_2$ so by inputting $P_1$, $T_1$, and $T_2$ you can calculate the final pressure. </p>
<p>EDIT: I do not believe you can easily calculate the time this takes.</p>
| 221
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thermodynamics
|
Amount of heat required at for unit rise in temperature at different temperatures of water
|
https://physics.stackexchange.com/questions/95527/amount-of-heat-required-at-for-unit-rise-in-temperature-at-different-temperature
|
<p>In my text book, it is given : </p>
<blockquote>
<p>One calorie is defined as the amount of heat required to raise the temperature of water from 14.5 °C to 15.5 °C.</p>
</blockquote>
<p>I found out in wikipedia that this is actually the definition of 15 °C calorie. </p>
<p>I want to know that as temperature increases, will the amount of energy for unit increase in temperature decrease? I find 15 °C calorie > 20 °C calorie. Is this relation uniform till 100 °C or is there any anomaly? And does the increased kinetic energy of the molecules of water, at an elevated temperature cause the amount of heat - required for unit rise in temperature, to be greater than the amount of heat required for the same unit rise, but at a lower temperature?</p>
<p>For example, the heat energy required to produce unit rise in temperature of water at 14.5 °C is 4.1855 joules and the amount of heat energy required to produce unit rise in temperature of water at 19.5 °C is 4.182 joules.</p>
<p>So does the increased K.E of water molecules at 19.5 °C cause the required heat energy to be lower than the heat energy required at 14.4 °C?</p>
|
<p>From the data <a href="http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html" rel="nofollow noreferrer">presented here</a>:</p>
<p><img src="https://i.sstatic.net/rxj1z.gif" alt="Cp of water"></p>
<p>Note that although the scale of the $y$ axis makes the variation in $C_p$ look big it is only about 1% over the range 0 to 100°C.</p>
| 222
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thermodynamics
|
Mixing of Ideal gas - Thermodynamic equilibrium
|
https://physics.stackexchange.com/questions/96089/mixing-of-ideal-gas-thermodynamic-equilibrium
|
<p>I always get confused what exactly happens when two ideal gases mix.</p>
<p>Consider the initial situation where two gases are in a box, separated by a rigid and adiabatic wall between them. Now when the wall between them is removed, they come to equilibrium (of course assuming the process is done quasi-statically). Initially the thermodynamic quantities of the gas be </p>
<p>$$ U_i,\: S_i,\:T_i,\:P_i,V_i \:\:\: \:\:\: \:\:\: \:\:\: \:\:\: \:\:\: \:\:\:i=1,2 $$
Now after the wall is removed, the condition for equilibrium is attained by the condition $$ dS = 0 $$
I am not able to apply this and find out exactly the final values of the thermodynamic quantities for an arbitrary ideal gas situation (I am not particularly able convince the situation physicaly)</p>
<p>(for instance if we consider and the two ideal gases to obey equation, $$ P_iV_i = c_iN_iRT_i \:\:\: \:\:\: \:\:\:i=1,2 $$
where $ c_i $ is the degrees of freedom of the gas, like $ c_i = \frac{3}{2},\:\frac{5}{2} $ for monatomic and diatomic respectively.</p>
|
<p>The total initial internal energy is
$U=U_1 + U_2=\frac{\nu_1}{2}c_1RT_1 + \frac{\nu_2}{2}c_2 RT_2 $
where the last equality comes from Joules' first law for ideal gases and where $c_i$ is the number of moles of species $i$ and $\nu_i$ is the number of degrees of freedom of the molecule (3 for atoms, 5 for diatomic molecules etc..).</p>
<p>Now, once equilibrium is reached everybody should have the same temperature $T$.
Since you are dealing with an ideal gas it implies that:</p>
<p>$$U= \frac{(\nu_1c_1+\nu_2c_2)RT}{2}$$ and hence since the whole system is isolated</p>
<p>$$T = \frac{\nu_1c_1T_1 + \nu_2c_2 T_2}{\nu_1c_1+\nu_2c_2}$$</p>
<p>Once the temperature is known, the rest follows easily. The pressure can be gotten straightforwardly as</p>
<p>$$P=\frac{(c_1+c_2)RT}{V_1+V_2}$$</p>
<p>because the ideal gas law is independent of the number of degrees of freedom of the different species.</p>
| 223
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thermodynamics
|
Does the molten salt used at projects such as Gemasolar need to be replaced after some cycles?
|
https://physics.stackexchange.com/questions/32018/does-the-molten-salt-used-at-projects-such-as-gemasolar-need-to-be-replaced-afte
|
<p>The project at <a href="http://www.torresolenergy.com/EPORTAL_DOCS/GENERAL/SENERV2/DOC-cw4cb709fe34477/GEMASOLARPLANT.pdf" rel="nofollow">Gemasolar</a> uses molten salt to store heat. This heat is then used with water to drive steam turbines. Over time does this kind of repeated heat/cold cycle make the molten salt a less effective medium to store heat? How would the old medium/waste be disposed of?</p>
|
<p>Gemasolar uses a mixture of sodium nitrate and potassium nitrate. There's an overview of this and other molten salt mixtures in <a href="http://files.halotechnics.com/Molten_salt_heat_transfer_fluid_with_low_melting_point-2010.Halotechnics.pdf" rel="nofollow">this pdf</a>.</p>
<p>The sodium/potassium nitrate is chosen because it has excellent stability, both chemical and thermal. The molten salt mixture will in principle last forever and wouldn't need to be replaced. Presumably the Gemasolar salt circulation system isn't perfect and will gradually build up impurities in the salt e.g. water from leaks or metal from wear in the pumps. How fast this will happen I don't know, and I suspect Gemasolar won't know either until they get experience at running the plant.</p>
| 224
|
thermodynamics
|
Multi-component mixture phase equilibrium criteria
|
https://physics.stackexchange.com/questions/33499/multi-component-mixture-phase-equilibrium-criteria
|
<p>For a pure species, the equlibrium between liquid phase and vapour phase is given by the equality of molar Gibbs energy in both phase:</p>
<p>$$\underline{G}^l=\underline{G}^v$$</p>
<p>Where $\underline{G}$ with an underline represents molar Gibbs energy.</p>
<p>In a multicomponent system, the equilibrium criterion is the equality of the <strong>partial</strong> molar Gibbs energy (aka chemical potential) for each component between each phases:</p>
<p>$$\overline{G}^l_i=\overline{G}^v_i$$</p>
<p>Now, because of the relation between molar quantities and partial molar quantities, $\underline{G}=\Sigma x_i\overline{G}_i$, the second criterion contains the first by setting $x_i=1$.</p>
<p>My question is : is the first criterion still valid between phases? If yes this would seem to imply that $x_i^l=x_i^v$, which is clearly wrong.</p>
|
<p>Your first criterion $ \underline{G}^l=\underline{G}^v $ does not apply for a mixture, though possibly not for the reasons you think.</p>
<p>With a pure material "a mole" has an unambiguous meaning, but for a mixture you have to choose what you're taking to be a mole. You are taking a weighted average of the components present, but unless the composition of the liquid and vapour phases are the same (in general they won't be) "a mole" of the liquid is not the same as "a mole" of the vapour. That's why your equation doesn't apply.</p>
<p>The only conditions under which "a mole" would be the same for the liquid and vapour phases is if the compositions of the liquid and vapour phases were the same, i.e. $x_i^l=x_i^v$. That's why you've ended up concluding that your criterion could only apply if $x_i^l=x_i^v$, and as you say that's usually wrong.</p>
| 225
|
thermodynamics
|
Do I have the meaning of the property temperature correct?
|
https://physics.stackexchange.com/questions/35898/do-i-have-the-meaning-of-the-property-temperature-correct
|
<p>OKay my book just starts out talking about the vague definition we have for temperature and we ended up with the Zeroth law of Thermodynamics which states:</p>
<blockquote>
<p>Two systems are in thermal equilibrium if and only if they have the same
temperature.</p>
</blockquote>
<p>So does that mean, the (brief) meaning of the property <em>temperature</em> is <strong>a measure of whether one system is in thermal equilibrium with another</strong>?</p>
|
<p>The answer is yes, as Alexander comments, but one should say that it is a little surprising--- it is saying that two systems, no matter their internal dynamics, will exchange heat according to a single real valued parameter. So that if you adjust this one parameter to be equal, then the systems will not exchange energy when in contact.</p>
<p>This can be stated axiomatically, if A doesn't exchange heat with B, and B doesn't exchange heat with C, then A doesn't exchange heat with C, also A doesn't exchange heat with a copy of A. These axiomatically define equivalence classes of systems "at the same temperature".</p>
<p>Further, if A is hotter than B (so energy flows from A to B on average) and B is hotter than C, then A is hotter than C (and this is a relation on equivalence classes). For any two systems A,B with A hotter than B, there is a C which is hotter than B and less hot than A. This tells you that temperature is linearly ordered.</p>
<p>All these axiomatic properties are a-priori surprising, since I didn't say anything about the Hamiltonian of the systems involved. They become obvious once you consider that systems maximize their phase-space volume (entropy). Then the temperature is defined statistically as the reciprocal of the rate of change of entropy with energy, so you always increase entropy by flowing heat from hot to cold. This explains why temperature is a linearly ordered real-valued equivalence-class type thing from simple first principles.</p>
| 226
|
thermodynamics
|
Gas pressure equalisation: Where does the excess energy go?
|
https://physics.stackexchange.com/questions/38839/gas-pressure-equalisation-where-does-the-excess-energy-go
|
<p>Assume for purposes of discussion a closed container with a concentration of gas on one side and near-vacuum on the other. We could let the gas pressure equalise naturally, or we could construct a barrier with a turbine such that the gas moving from the concentrated side to the vacuum side spins the turbine and produces electricity. In both cases, the gas reaches equilibrium after some time, but in one of the cases the system performs work. My question is, what happens to the excess energy <em>not</em> used to produce electricity in the first example (with no turbine).</p>
<p>My intuition tells me that the answer has to be one of the following: Either the gas that drove the turbine is colder than the gas that did not, or that the electrical energy produced is actually less than the energy required to build the turbine so the electrical energy was actually input into the system by action of building the turbine.</p>
<p>Is either of my answer right? What else have I not accounted for?</p>
<p>Thanks.</p>
|
<p>Zephyr's comment answers your question (Zephyr if you want to post it as an answer I'll delete this) but I think it's worth expanding (no pun intended :-) on it a bit.</p>
<p>Consider the microscopic view: assuming an ideal gas the internal energy is just the total kinetic energy of the gas molecules, and the temperature is a measure of their speed.</p>
<p>If you sudden remove the barrier in your system and let the gas expand freely the speed and kinetic energy of the gas molecules doesn't change. However if you have a turbine then gas molecules will hit the turbine blades and transfer some of their energy to the turbine. That means their speed and therefore temperature will decrease. If you were to measure the total kinetic energy of the gas molecules after expansion through the turbine you would find it had decreased by an amount equal to the energy generated by the turbine.</p>
| 227
|
thermodynamics
|
How to evaluate heating value on the basis on these data?
|
https://physics.stackexchange.com/questions/39033/how-to-evaluate-heating-value-on-the-basis-on-these-data
|
<p>Heating value of liquid-gas mixture was tested with primus which has gas burner. In the initial state aluminum container with lid protected from the wind has temperature of 0 celsius and in it 0,54 kg of water and 0,82 kg of ice cubes. Mixture was then heated to 100 celsius. During the heating 0,022 kg of liguid-gas was consumed. Evaluate heating value on the basis on these data. What factors are likely to contribute to the fact that the heating value that you get diverges from the real Heating value? Just some hint please I think I could use equation of thermal energy $U=C(T)\Delta T$, but then what you need the masses of water and ice cubes for?
I could not find any formulas for heating value, why?</p>
|
<p>The reason your professor has specified a mixture of ice and water is to check that you know about latent heat of fusion. Your professor won't be amused if I say any more, but a quick Google should tell you what extra bit you need to do when calculating the heat required to get the water and ice to 100C. If you're still unsure please ask as a comment to this answer.</p>
| 228
|
thermodynamics
|
Thermal Expansion. Is there a way to fix my bottle?
|
https://physics.stackexchange.com/questions/38529/thermal-expansion-is-there-a-way-to-fix-my-bottle
|
<p>I did something very stupid tonight. I poured hot water into an polyethylene water bottle and it immediately contracted causing it to overflow and burn my hand and leg. I immediately wasted my hand and feet with cold water.</p>
<p>Now I am left with a useless bottle that resembles what you see at a modern art fair. Is there a way to thermally recover my bottle?</p>
<p>How could I 'thermally' expand it into it's natural shape again?</p>
|
<p>Your bottle was probably made from high density polyethylene. When you heat this this the molecules within it rearrange and become less crystalline, and this causes the shrinkage. In principle you could chop up the bottle and remould it, but nothing short of this is going to recover it. Sorry!</p>
| 229
|
thermodynamics
|
Is it true that the mass of air in normal atmospheric pressure over 1$m^2$ is 10326 $kg$?
|
https://physics.stackexchange.com/questions/39553/is-it-true-that-the-mass-of-air-in-normal-atmospheric-pressure-over-1m2-is-10
|
<p>Is it true that the mass of air in normal atmospheric pressure over 1$m^2$ is 10326 $kg$?
I calculated it from pressure formula </p>
<p>$p=\frac{F}{A}$. </p>
<p>Let $m=?, A=1 m^2 and p=101300 Pa$. </p>
<p>$p=\frac{F}{A} \Leftrightarrow ...\Leftrightarrow
m= \frac{pA}{g}=\frac{101300 \cdot 1 }{9,81} = 10326 kg $.
So why we don't feel that weight, because if $A=160 cm^2 = 160 \cdot 10^{-4} m^2$( the area of normal man
s feet), then
$m=\frac{pA}{g}=\frac{101300 \cdot 160 \cdot 10^{-4} }{9,81} =165 kg$.</p>
<p>Am I missing something? Why we don't feel that weight?</p>
|
<p>Your calculations are correct. The reason we don't feel it is because the 1 atmosphere of pressure will be applied to all surfaces of our body, including the soles of our feet. In fact the interior of our body is also pushing out against our skin with 1 atmosphere of pressure so there is no net force being applied in either direction at the skin's surface. The only way you could feel this pressure would be if part of your body was in contact with a vacuum while the rest of your body remained in air at 1 atmosphere of pressure. Imagine that you were standing on an opening to a vacuum chamber that was almost, but slightly less than the size of the soles of your feet. If there was a perfectly air tight fit between the vacuum opening and your feet, you would indeed feel that pressure as that much force on your feet just as you calculated. As a simpler experiment, just put your hand over the hose of a vacuum cleaner - that is only a very partial vacuum but the force you need to exert to remove your hand is due to the 1 atmosphere of pressure around us all pushing your hand onto the partial vacuum. In fact if you measured that force needed to pull your hand off of the vacuum cleaner hose, you could compute what the air pressure in the vacuum hose was.</p>
| 230
|
thermodynamics
|
What's the lower limit for energy usage to desalinate water?
|
https://physics.stackexchange.com/questions/43955/whats-the-lower-limit-for-energy-usage-to-desalinate-water
|
<p>Consider a desalination process where you enter sea water and receive fresh water and brine (or maybe pure salt).</p>
<p>How do I compute the least amount of energy per mass? I think this has something to do with the entropy of the different mixtures, mostly dependent on the salt concentration in the brine.
Is this correct? How to calculate?</p>
|
<p>Assuming that the process is done at constant temperature and pressure, what you want is the Gibbs free energy difference. The reaction could be taken to be 1 mole of water in brine on one side and one mole of pure water on the other side.</p>
<p>The result should be a positive number since the process is not spontaneous. However the value of $\Delta G$ will be the minimum non-mechanical work necessary for the process.</p>
<p>To understand this read up on the Gibbs free energy.</p>
| 231
|
thermodynamics
|
What is the importance of Joule's experiment?
|
https://physics.stackexchange.com/questions/45212/what-is-the-importance-of-joules-experiment
|
<p>I was reading about <a href="http://it.wikipedia.org/wiki/Mulinello_di_Joule" rel="nofollow">the experiment of Joule</a> (Italian wiki page). I'm not sure how it's called in English, since there is only an Italian and French version. In any case, in the page it is stated that (my translation):</p>
<blockquote>
<p>Through this experiment, Joule determined the <a href="http://en.wikipedia.org/wiki/Mechanical_equivalent_of_heat" rel="nofollow">mechanical equivalent of heat</a> to be equal to 4,186 J/cal, value of extraordinary precision for the time.</p>
</blockquote>
<p>Is the importance of this experiment "only" ascribable to the fact that the evaluation was really precise and exact for the time? Or are there other reasons? </p>
<p>I don't need a whole list of reasons, just the most exemplary/important<sup>1</sup> fact(s) will suffice. </p>
<hr>
<p><sup>1: Objectively considered as such by most researchers, experts, etc.</sup></p>
|
<p>The answer can be found in the <a href="http://en.wikipedia.org/wiki/Mechanical_equivalent_of_heat" rel="nofollow">Wikipedia page</a> you linked to! Historically, heat had been considered a substance, called <a href="http://en.wikipedia.org/wiki/Caloric_theory" rel="nofollow">caloric</a>. Joule's experiment proved that heat was actually a form of mechanical energy, so was a crucial step towards our modern understanding of the <strong>conservation of energy</strong>.</p>
| 232
|
thermodynamics
|
Calculating the $\frac{dp}{dT}$ slope using Clausis-Clapeyron
|
https://physics.stackexchange.com/questions/48302/calculating-the-fracdpdt-slope-using-clausis-clapeyron
|
<p>I've produced experimental data over how the boiling point of water varies with pressure and temperature and plotted this in a PT graph. I would like to verify my results using theory. The <a href="http://en.wikipedia.org/wiki/Clausius-Clapeyron_relation" rel="nofollow">Clausius-Clapeyron equation</a> appears to be exactly what I want. I did manage to find a table over <a href="http://en.wikipedia.org/wiki/Properties_of_water#Heat_capacity_and_heats_of_vaporization_and_fusion" rel="nofollow">heat of vaporization</a> depending on pressure, so that this version of Clausius-Clapeyron is almost applicable:</p>
<p>$$
\frac{dP}{dT}=\frac{L}{T\Delta v}
$$</p>
<p>However, I don't have any values for $\Delta v$. So what are my options? Is there some approximation I can make to find the value for $\Delta v$ or is there another version of Clausius-Clapeyron I can use to find $\frac{dP}{dT}$ using freely available tables? Or would someone suggest another way of verifying my results?</p>
|
<p>The usual approximation is to disregard liquid volume against vapor volume, and to consider the latter to be that of an ideal gas, so you get $\Delta v = RT - 0 = RT$ and so</p>
<p>$$
\frac{dP}{dT}=\frac{L}{RT^2}
$$</p>
<p>If you simply want to check your results, there are also plenty of available resources on-line with boiling points for water as a function of pressure, such as <a href="http://www.engineeringtoolbox.com/boiling-point-water-d_926.html" rel="nofollow">this</a>.</p>
| 233
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thermodynamics
|
heating verses air conditioning
|
https://physics.stackexchange.com/questions/51739/heating-verses-air-conditioning
|
<p>My Co-workers and I are trying to figure this out, but can't think of a logical answer. It's probably an easy one, but I'll ask anyway-
Ok, the heater is set at 73 degrees.
The a/c is set at 73 degrees.
Why does it feel warmer with the heat on if they're both 73 degrees?</p>
|
<p>1) Maybe it's your imagination<br>
2) When the heater is on, its cold outside, so maybe it feels relatively warmer inside (and conversely for when the A/C is on)<br>
3) Maybe the cold air from the A/C has more of a chilling effect than the warm air from the heater has a warming effect, thus making it feel colder with the A/C<br>
4) Maybe something about the aerodynamics in your building i) increase the draftiness when the A/C is on (or its warmer outside), or ii) cause the thermostat to measure a lower temperature when the heat is on<br>
5) Maybe the system is designed to over-heat more than it over-cools<br>
6) Maybe something else</p>
| 234
|
thermodynamics
|
The physical laws of thermodynamics tell us that electrical impulses must produce heat?
|
https://physics.stackexchange.com/questions/53447/the-physical-laws-of-thermodynamics-tell-us-that-electrical-impulses-must-produc
|
<blockquote>
<p>Health - Controversial New Idea: Nerves Transmit Sound, Not Electricity - By Robert Roy Britt, LiveScience Managing Editor
posted: 14 March, 2007 1:00 pm ET Rendering shows a biological membrane at its melting point. The green molecules are liquid, and the red are solid. Molecules of anesthetic reduce the number of red areas so that the sound pulse can no longer transport its signal. The nerve is anesthetized. Credit: Heiko Seeger, Ph.D. Niels Bohr Institute
Nerves transmit sound waves through your body, not electrical pulses, according to a controversial new study that tries to explain the longstanding mystery of how anesthetics work.
Textbooks say nerves use electrical impulses to transmit signals from the brain to the point of action, be it to wag a finger or blink an eye.
"But for us as physicists, this cannot be the explanation," says Thomas Heimburg, a Copenhagen University researcher whose expertise is in the intersection of biology and physics. "The physical laws of thermodynamics tell us that electrical impulses must produce heat as they travel along the nerve, but experiments find that no such heat is produced."</p>
</blockquote>
<p>How should I read:</p>
<blockquote>
<p>"The physical laws of thermodynamics tell us that electrical impulses <strong>must produce heat</strong> as they travel along the nerve, but experiments find that no such heat is produced." ?</p>
</blockquote>
<p>I learned: </p>
<p>If only resistive [being coupled to an electrical resistance] electrical signal transmission takes place, there is dissipation of electric energy in the relevant resistance and hence development of heat.</p>
<p>But also, I learned:</p>
<p>If the signal is purely capacitive [a circuit of ideal capacitors] or inductively [ideal coils], no energy is dissipated.</p>
<p>Under ideal capacitors and inductors current and voltage differ always 90 ° in phase. </p>
<p>Therefore φ = 90 °.</p>
<p>If the current <em>I</em> and the voltage <em>V</em>, then the energy dissipation is always given by:
<em>I</em> × <em>V</em> × cosφ</p>
<p>And because cos 90 ° = 0, is also the dissipation 0.</p>
<p>Is "purely capacitive" or "inductively" not the explanation why no such heat is produced?</p>
|
<p>The quote you give is very vague and hand waving. Various theories have been proposed of how nerve cells transmit information, which one can find by googling. The <a href="http://en.wikipedia.org/wiki/Synapse" rel="nofollow">models</a> are complex and this seems to be one more in the row.</p>
<p>You are correct in that inductive and capacitive transfers would not heat the nerve.
It seems that they expect the signal transfer if electric to be resistive.
It is not an extreme assumption because it is hard to model capacitive or inductive transfer in the messy environment of a nerve.</p>
<p>One needs more information to see whether their assumption of sound transfers is reasonable or not. Others have been proposing a type of superconductivity to explain nerve message transfers.</p>
| 235
|
thermodynamics
|
What happens to the time it takes for my noodles to cook, and the temperature of the boiling water?
|
https://physics.stackexchange.com/questions/54709/what-happens-to-the-time-it-takes-for-my-noodles-to-cook-and-the-temperature-of
|
<p>I have a boiling pot of water that I am going to cook my Sponge Bob Noodles in. I am very hungry and want the noodles to cook faster, so I turn up the heat from medium to high.( Note, the water was already boiling at medium).What happens to the time it takes for my noodles to cook, and the temperature of the boiling water?</p>
|
<p>When you put in the noodles the temperature of the water will fall below the boiling point for a time interval delta(t) until it reaches boiling point again, i.e. the noodles go from room temperature to boiling point.</p>
<p>This interval becomes smaller the higher the energy provided during the transition period. So if you are in a hurry, provide high heat. You might gain a minute or three, depending on the heat source and pot and volume of water and noodles. </p>
<p>If you are energy conscious turn down the heat after it boils again.</p>
| 236
|
thermodynamics
|
At what depth in the water atmospheric pressure is 100 times greater than on the ground?
|
https://physics.stackexchange.com/questions/53758/at-what-depth-in-the-water-atmospheric-pressure-is-100-times-greater-than-on-the
|
<p>At what depth in the water atmospheric pressure is 100 times greater than on the ground?
This question comes from the fact that average pressure in Earth( 1000 mbar) is 100 times greater than in Mars( 7 mbar).</p>
|
<p>The atmospheric pressure at STP is 101325 N/m$^2$, so 100 times this is 1.01325 $\times$ 10$^7$ N/m$^2$. You just have to work out the height of a column of water with a 1 m$^2$ base and weighing 1.01325 $\times$ 10$^7 /g$ kg, where $g$ is the acceleration due to gravity. I make it about 1.03 kilometers, though note that it will vary slightly with temperature because the density of water varies with temperature.</p>
| 237
|
thermodynamics
|
Most efficient type of heat pump
|
https://physics.stackexchange.com/questions/56576/most-efficient-type-of-heat-pump
|
<p>Can I make a heat pump beat the Carnot efficiency? Why is the <a href="http://en.wikipedia.org/wiki/Carnot_cycle" rel="nofollow">Carnot process</a> the most efficient one?</p>
<p>If I have a heatpump that is sphere shaped, and cascaded in layers like a onion can I beat Carnot efficiency? Heat would be transferred and concentrated from the outermost layer that is in contact with ambient air, and brought towards the center to heat a fluid.</p>
|
<p><strong>No.</strong></p>
<p>The laws of thermodynamics have nothing to do with specific engineering designs - they state properties of the time-evolution of certain observables of physical systems, <em>independent</em> of design. Nor are they empirical laws - they are derived rigorously from first principles that no one doubts.</p>
<p>You can no more beat the Carnot efficiency by clever construction than you can build a time machine if only you had enough gears. Just as no amount of pistons and lasers could ever change the rest mass of the electron, no amount of Peltier cells can undo physics.</p>
<p>Thousands of people have thought along similar lines - "maybe if I just tweak things, my design will work." When they actually do build their physically impossible device and claim it works, the fundamental flaw is always one of two varieties: (1) improper calorimetry misses sources or losses of energy, or (2) the temperature reservoirs are not really at the temperatures the inventors suspect. Without more detailed design specs, I can't say where your reasoning fails, but doing that analysis is beyond the scope of physics anyway.</p>
| 238
|
thermodynamics
|
How does a thermal temperature gun work?
|
https://physics.stackexchange.com/questions/60115/how-does-a-thermal-temperature-gun-work
|
<p>I once worked as a kitchen porter over a winter season.<br>
We had fun with thermal temperature guns (like <a href="http://www.amazon.co.uk/tag/infrared%20thermometer/products" rel="nofollow">these</a>) which I learned can be used for measuring the temperature of something a reasonable distance away (aside from the obvious use of <a href="http://en.wikipedia.org/wiki/Laser_tag" rel="nofollow">laser tag</a>), which to my mind is pretty impressive.</p>
<p>How do they work?</p>
|
<p>They basically measure the intensity of the infrared blackbody radiation in some wavelength region and calculate the temperature needed to give that intensity according to <a href="http://en.wikipedia.org/wiki/Planck%27s_law" rel="nofollow">Planck's law</a>.</p>
| 239
|
thermodynamics
|
Computer cooling with dry ice, ideas and question; thermodynamics
|
https://physics.stackexchange.com/questions/60883/computer-cooling-with-dry-ice-ideas-and-question-thermodynamics
|
<p>I am designing a cooling system for my computer and had a few questions.</p>
<p>So I have a computer water cooling radiator and I want to cool it as much as I can.</p>
<p>My first two ideas were an old window A/C unit and blowing air over the cold radiator in there onto my radiator. Or put the whole radiator again in water and fill the water with ice indefinetely...</p>
<p>My next idea, which I am going to pursue, is to set dry ice on the radiator. Or to submerge the radiator in some sort of liquid with a low freezing point (acetone?) and get that liquid as cold as I can by sublimation the dry ice in it. </p>
<p>Using the acetone is a good idea because it can grab all the cold from the dry ice (surface area is covered by liquid; opposed to it just sitting on the rad and 3/4 of the dry ice is incontact with the air doing nothing) But I am not sure how well my radiator would stand up to being submerged in acetone (not to mention explosions...)</p>
<p>So what would be my best bet to using the dry ice to chill the water? How do I calculate how much I need for the amount of watts in the watter?</p>
<p>I know my water cooling loop must remove about 625watts at max computing power, so how much dry ice do I need to remove that amount of heat?</p>
<p>I think setting dry ice on it works well (have done it before) but I am scared of local freezing inside the radiator, and that ice hitting my pump impellor then (no-no). Also if I get the water too cold I will start to get condensation on some parts (no-no). </p>
|
<p>A safer solution - buy everclear. </p>
<p>1) Mix one part ethanol to three parts water. The mixture freezes at around -15C. Make ice cubes out of most of this mixture, cool the rest. Your freezer should be able to just freeze this ethanol+water mixture, if not add some water. </p>
<p>2) Replace the water in your computer cooler with a 30% ethanol - 70% water mixture. This won't freeze and jam your impeller, and it wont burn (the rubber might dissolve though) </p>
<p>3) Immerse your radiator into a bucket of full of chilled 25% ethanol and pour the ice cubes into it. </p>
<p>4) As your ice cubes melt, replenish them. As the ethanol evaporates out occasionally add a shot of ever-clear. </p>
<p>Now you switched the unpleasantness of CO2 intoxication, for the pleasure of ethanol intoxication! </p>
<p>(As noted in the comments, with such low temperatures you'll have condensation problems. So buy a bunch of silica packets and dump them into your case and seal it up. Or blow dry hot air on exposed electronic parts. Or whatever)</p>
| 240
|
thermodynamics
|
phase transition by sublimation
|
https://physics.stackexchange.com/questions/61801/phase-transition-by-sublimation
|
<p>Why do sublime solids sublimate in the first place? Is it that their melting point and their liquid's boiling point are the same so that they manage to magically skip the liquid phase? (if so, why does it have to be so?)And is the reverse possible?</p>
|
<p>$CO_2$ sublimates at atmospheric pressure, lets have a look at the phase diagram.</p>
<p><img src="https://i.sstatic.net/wH8wt.jpg" alt="Phase diagram with triple point"></p>
<p>Now if you notice the triple point exists above atmospheric pressure. Below this the system can only transform directly from the solid to the gas phase and vice versa by changing the temperature. At the triple point the substance can coexist in all three phases at one distinct temperature and pressure. Beyond the triple point the three phases are distinct with respect to temperature, so via heating or cooling you can witness all three different states.</p>
<p>The opposite of sublimation exists and is called deposition, an example of which would be frost.</p>
<p>A link to a deeper explanation:
<a href="http://www.av8n.com/physics/melt-sublimate.htm" rel="nofollow noreferrer">http://www.av8n.com/physics/melt-sublimate.htm</a></p>
| 241
|
thermodynamics
|
Finding equation of state from thermal expansion coefficient and isothermal compressibility
|
https://physics.stackexchange.com/questions/62293/finding-equation-of-state-from-thermal-expansion-coefficient-and-isothermal-comp
|
<p>I'm stuck on a problem that I found in a book (Modern Thermodynamic with Statistical Mechanics, Helrich S., problem 5.2).</p>
<p>The text of the problem is that:</p>
<blockquote>
<p>Consider a solid material for which:</p>
<p>$$
\frac{1}{\kappa_T} = \frac{\varepsilon}{2V_0}\left[\frac{2\Gamma c_v T}{\varepsilon}\,\frac{V_0}{V} - 3\left(\frac{V_0}{V}\right)^3\right]
$$
$$
\beta = \frac{1}{T}\left[1 + 3\,\frac{\varepsilon}{2\Gamma c_v T}\left(\frac{V_0}{V}\right)^2\right]
$$</p>
<p>Where $\varepsilon$ is a constant with the units of energy, $\Gamma$ is a dimensionless constant and $V_0$ is a reference volume less than $V$. The temperature range is such that we may assume that the specific heat at constant volume $c_v$ is independent of temperature.
Find the thermal equation of state.</p>
</blockquote>
<p>In this book the convention defines $\kappa_T$ as the isothermal compressibility and $\beta$ as the thermal expansion coefficient.</p>
<p>$$
\beta = \frac 1V \left(\frac{\partial V}{\partial T}\right)_P
$$</p>
<p>$$
\kappa = - \frac 1V \left(\frac{\partial V}{\partial P}\right)_T
$$
The answer key for this problem says:</p>
<p>$$
P(V,T) = \frac{\Gamma c_vT}{V} + \frac{\varepsilon}{2V_0}\left[\left(\frac{V_0}{V}\right)^5 - \left(\frac{V_0}{V}\right)^3\right]
$$</p>
<p>Here's the procedure I tried to apply.
I wanted to do a contour integration of this equation because I can express the partial derivatives as a function of $\kappa _T$ and $\beta$:
$$
dP(T,V) = \left(\frac{\partial P}{\partial V}\right)_TdV + \left(\frac{\partial P}{\partial T}\right)_VdT
$$
given that:
$$
\left(\frac{\partial P}{\partial T}\right)_V = \frac{\beta}{\kappa_T}
$$
$$
\left(\frac{\partial P}{\partial V}\right)_T = - \frac{1}{V \kappa_T}
$$</p>
<p>But by using this procedure I have a problem with the integration of $c_v$ with respect to the volume, and even if I assume this constant I still don't get the result.</p>
<p>Which approach would you suggest? </p>
|
<p>Your approach is all right but the solution given by the textbook is wrong :), at least if no approximation is to be made.</p>
<p>Let's go the other way around: start from $$P(V,T) = \frac{\Gamma c_vT}{V} + \frac{\varepsilon}{2V_0}\left[\left(\frac{V_0}{V}\right)^5 - \left(\frac{V_0}{V}\right)^3\right]$$ and then derive the values of $\kappa_T$ and $\beta$ (by the way, shouldn't it rather be $\chi_T$ (<a href="http://fr.wikipedia.org/wiki/Compressibilit%C3%A9" rel="nofollow">see here</a>) and $\alpha$ (<a href="http://en.wikipedia.org/wiki/Thermal_expansion" rel="nofollow">here</a>)?). As you mentioned, to do this, we need to compute partial derivatives of $P$:</p>
<p>$$
\left(\frac{\partial P}{\partial T}\right)_V = \frac{\Gamma c_v}{V}
$$</p>
<p>$$
\left(\frac{\partial P}{\partial V}\right)_T =
-\left(\frac{\Gamma c_vT}{V^2} + \frac{\varepsilon}{2{V_0}^2}\left[5\left(\frac{V_0}{V}\right)^6 - 3\left(\frac{V_0}{V}\right)^4\right]\right)
$$</p>
<p>which give</p>
<p>$$
\frac{1}{\kappa_T} = -V \left(\frac{\partial P}{\partial V}\right)_T
= \frac{\Gamma c_vT}{V} + \frac{\varepsilon}{2{V_0}}\left[5\left(\frac{V_0}{V}\right)^5 - 3\left(\frac{V_0}{V}\right)^3\right]
$$
and</p>
<p>$$
\beta = \kappa_T \left(\frac{\partial P}{\partial T}\right)_V
= \frac1T \left(1 + \frac{\varepsilon }{2\Gamma c_V T}\left[5\left(\frac{V_0}{V}\right)^4 - 3\left(\frac{V_0}{V}\right)^2\right]
\right)^{-1}
$$</p>
<p>Clearly, that's not what the textbook gives in the first place, but it's close enough to understand what they did: a Taylor expansion at order 3 in $V_0/V$ in booth cases, which gives back the expressions</p>
<p>$$
\frac{1}{\kappa_T}
\approx \frac{\Gamma c_vT}{V} + \frac{\varepsilon}{2{V_0}}\left[ - 3\left(\frac{V_0}{V}\right)^3\right]
$$</p>
<p>and</p>
<p>$$
\beta
\approx \frac1T \left(1 + \frac{\varepsilon }{2\Gamma c_V T}\left[ - 3\left(\frac{V_0}{V}\right)^2\right]
\right)^{-1}
\approx \frac1T \left(1 + \frac{\varepsilon }{2\Gamma c_V T}\left[ 3\left(\frac{V_0}{V}\right)^2\right]
\right)
$$</p>
<p>It really should have been clearer in the text that you could suppose $V\gg V_0$ (and not only $V>V_0$) and I don't see how you could derive the term in $\left(\frac{V_0}{V}\right)^5$ from this as it is completely neglected to find back $\kappa_T$ and $\beta$</p>
| 242
|
thermodynamics
|
What is the work done by an ideal gas?
|
https://physics.stackexchange.com/questions/62792/what-is-the-work-done-by-an-ideal-gas
|
<p>What is the work done by an ideal gas when final pressure and volume are both different from its initial pressure and volume or when both pressure and volume changes ?</p>
|
<p>The work done by the gas on the environment is $W = \int_{V_a}^{V_b} P(V)\text{d}V$, where $V_a$ and $V_b$ are the initial and final volumes of the gas and $P(V)$ is the pressure exerted by the gas on the environment when the gas has volume $V$.</p>
<p>There is no simpler expression unless you specify something else about $P$. Work is a path-dependent quantity; there is no way to express the work as a function of simply the starting and ending states of the system.</p>
| 243
|
thermodynamics
|
Error in Sear's and Zemansky's University Physics with Modern Physics 13th Edition (Young and Freeman)?
|
https://physics.stackexchange.com/questions/64026/error-in-sears-and-zemanskys-university-physics-with-modern-physics-13th-editi
|
<p>I was reading up on the Ideal Gas Equation in University Physics with Modern Physics by Young and Freeman when I chanced upon a seemingly illogical mathematical equation.</p>
<p>Can anyone rectify this error? Or is it misunderstanding on my part?</p>
<p>Here is the portion (Page 600, Chapter 18, Equation 18.12):</p>
<blockquote>
<p>$$pV = \frac{1}{2}Nm(v^2)_{av} = \frac{1}{3}N\biggl[\frac{1}{2}m(v^2)_{av}\biggr]$$</p>
</blockquote>
<p>It should be clear that $\frac{1}{3} \neq (\frac{1}{3} \times \frac{1}{2})$.</p>
|
<h2>Nice catch!</h2>
<p>For reference here is the book page.</p>
<p>:<img src="https://i.sstatic.net/TnAnI.png" alt="enter image description here"></p>
<p>See , though it may error in printing or anything else.The final equation they get $$pV=\dfrac23K_{tr}$$ is very correct.</p>
<p>The correct form of $eq.(18.12)$ must be $$pV=\dfrac13Nm(v^2)_{av}=\dfrac
{\color{red}{\huge{2}}}3N\bigg[\dfrac12 m (v^2)_{av}\bigg]$$</p>
| 244
|
thermodynamics
|
Explain to me how this water cooler works?
|
https://physics.stackexchange.com/questions/64092/explain-to-me-how-this-water-cooler-works
|
<p>Goodday all, </p>
<p>I was recently reading up on a few projects that might be of interest to me when I found "CPU Bong water coolers", there isn't much online on them so I figure I would ask y'all.</p>
<p>If you do a quick google search you can see what this thing actually is. </p>
<p>They get a length of PVC and orient it vertially, at they bottom they attach a 45 split with the 45 open end facing up. The very bottom of the pipe sets in a bucket with water.</p>
<p>Water from the bucket is pumped through a CPU waterblock and collects heat. The water continues to a shower head situated at the top of the PVC pipe. The water flows through the shower head and strait down the tube into the bucket. While the water is falling through the tube air is forced over it via a fan over the open end of the 45 split. </p>
<p>I think the idea is that the air gonig over the small droplets of water will strip off molecules that are close to evaporation. When they evaporate they pull heat from the last of the water in that drop, cooling the drop. All the drops are cooled this way.</p>
<p>Is this how this thing works? </p>
<p>How could I get a rough estimate the size I would need to remove lets say 300 watts from the incoming water? (I know I will get people who quote "But what if the butterfly flaps its wings in Europe, that thing will then only cool 299.4553456 watts; I just need something generic. Should I get a 10foot PVC pipe or a 6in pipe?)</p>
<p>Does it scale? If a 1ft pipe can effectively remove 10 watts, does a 10ft pipe remove 100 watts?</p>
<p>If you think it depends on to many factors to tell me, what are the factors? Humidty, ambient, CFM of the fans, how small the drops are from the shower head, etc and which should be maximized and which should be minimized.</p>
| 245
|
|
thermodynamics
|
Specific heat and temperature
|
https://physics.stackexchange.com/questions/6367/specific-heat-and-temperature
|
<p>This is a two-part question...</p>
<p>Firstly, models of the specific heat capacity $C$ (i.e. Debye, Einstein) in relation to the temperature $T$ give $C$ as steadily increasing with $T$. I assume that the change in $C$ is due to the heat the system gains being stored in its degrees of freedom- so why doesn't it increase in steps?</p>
<p>Secondly, what value does the temperature refer to? If the heat supplied raises the temperature from $T_1$ to $T_2$, does $C$ refer to $T_1$, $T_2$ or somewhere in between? If, in experiment, a large temperature difference was used, can it be assumed that $C$ changes over the range and so the answer is an average/useless?</p>
|
<p>When you talk about the heat capacity increasing in steps, I assume you're referring to the equipartition principle, which says that the heat capacity is <span class="math-container">${1\over 2}k_B$</span> times the number of degrees of freedom. As more degrees of freedom become available, the heat capacity goes up in steps of <span class="math-container">${1\over 2}k_B$</span>.</p>
<p>The thing is that equipartition only applies in some circumstances, roughly speaking when the degrees of freedom "look classical," in the sense that the separation between energy levels is small compared to <span class="math-container">$k_BT$</span> (among other conditions). Those assumptions don't hold for the Debye and Einstien models. In particular, as the temperature is raised, more quantum levels become available, and those newly-available ones, more or less by definition, aren't in that classical regime.</p>
<p>For your second question, the heat capacity only applies to infinitesimal increments. That is, instead of writing
<span class="math-container">$$
Q=C\,\Delta T,
$$</span>
you have to write
<span class="math-container">$$
\delta Q=C\,dT.
$$</span>
if <span class="math-container">$C$</span> varies significantly over the temperature interval in question, you have to integrate
<span class="math-container">$$
\int_{T_1}^{T_2}C(T)\,dT
$$</span>
to get the required heat input.</p>
| 246
|
thermodynamics
|
time dependence of temperature equalization
|
https://physics.stackexchange.com/questions/7764/time-dependence-of-temperature-equalization
|
<p>Suppose you have two thermodynamical systems $X_1$ and $X_2$ (for example water and air) with different temperatures ($T_1 > T_2$). Now put them into thermal contact. Is there a formula which describes, how $T_1$ and $T_2$ changes with time?</p>
|
<p>In a simple heat conduction theory (no convection) the corresponding time is determined with the lowest eigenvalue of the Sturm-Liouville problem: $\Delta T(t) = A\cdot e^{-t/\tau}$ (regular regime), see, for example, my <strong><a href="http://arxiv.org/abs/0906.3504" rel="nofollow">article</a></strong>. </p>
<p>EDIT: For any time $t > 0$ the temperature difference is a series like</p>
<p>$\Delta T(t) = \sum_{n=0}^\infty A_n \cdot e^{-t/\tau_n}$</p>
<p>$\tau_{n+1} < \tau_n$, for example, $\tau_n \propto \frac{1}{\pi^2(n+1)^2}$.</p>
<p>When $t >> \tau_1$, the fast decaying exponentials $e^{-t/\tau_n}$ are small with respect to the slowest one $e^{-t/\tau_0}$ so only one term remains in this sum. The latter regime is called a regular regime of the heat exchange. For certain one-layer 1D systems $\tau \propto \frac{\rho c L^2}{\kappa \pi^2(n+1)^2}$ Here $\kappa$ is the heat conductivity, $\rho$ is the material density, $c$ is the specific heat capacity, and $L$ is the layer thickness (exact $n$-dependence is determined with the boundary conditions).</p>
| 247
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thermodynamics
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Can a water bubble be frozen into ice bubble?
|
https://physics.stackexchange.com/questions/7780/can-a-water-bubble-be-frozen-into-ice-bubble
|
<p>To maintain the surface tension which formed our original bubble (in order to keep the bubble from breaking), we may change the temperature/pressure of air on both sides of the bubble varyingly, with different rates.</p>
<p>Is such a configuration possible? Examples of any such system?</p>
|
<p>The answer is <a href="http://en.wikipedia.org/wiki/Soap_bubble#Freezing">YES</a>. Note that <a href="http://en.wikipedia.org/wiki/Surface_tension#Surfactants">pure water bubbles are unstable</a> but it should be possible to freeze one without it popping as long as it is thick enough and the pressure is correctly regulated. (<a href="http://www.youtube.com/watch?v=ddST_7n9peg">This video</a> shows in detail a soap bubble partially freezing and then bursting on contact, and <a href="http://www.youtube.com/watch?v=OM9WXrGftXE">this video</a> shows more completely frozen soap bubbles rolling around and eventually disintegrating.) If completely frozen, the bubble will no longer be a perfect sphere because the ice crystal will nucleate at certain positions and this will distort the surface.</p>
| 248
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thermodynamics
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Why is it when you microwave cold coffee and then add milk it creates a foam head?
|
https://physics.stackexchange.com/questions/8542/why-is-it-when-you-microwave-cold-coffee-and-then-add-milk-it-creates-a-foam-hea
|
<p>As compared to when the coffee is just hot from brewing. I suspect it has something to do with the way the microwaves are affecting the molecules of the coffee.</p>
| 249
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thermodynamics
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Does the Grand Canonical Ensemble allow for exchange of particles or not?
|
https://physics.stackexchange.com/questions/8714/does-the-grand-canonical-ensemble-allow-for-exchange-of-particles-or-not
|
<p>I was doing some reading on wikipedia and found it interesting that one page says the Grand Canonical Ensemble does not allow for exchange of particles, however another page says it does. So I went on google books and tried to look for a more trust worthy source, again the same happens one source says it allows the other says it doesn't so which is it?</p>
<p>Book that says it does allow exchange:
<a href="http://www.scribd.com/doc/52426748/46/Grand-Canonical-Ensemble" rel="nofollow">http://www.scribd.com/doc/52426748/46/Grand-Canonical-Ensemble</a></p>
<p>Another that says it isn't allowed:
<a href="http://books.google.co.uk/books?id=5sd9SAoRjgQC&pg=PA62&dq=grand+canonical+ensemble+exchange+particles&hl=en&ei=hL6oTZ2THc_p4wbEh-3DCg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CC0Q6AEwAA#v=onepage&q=grand%20canonical%20ensemble%20exchange%20particles&f=false" rel="nofollow">http://books.google.co.uk/books?id=5sd9SAoRjgQC&pg=PA62&dq=grand+canonical+ensemble+exchange+particles&hl=en&ei=hL6oTZ2THc_p4wbEh-3DCg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CC0Q6AEwAA#v=onepage&q=grand%20canonical%20ensemble%20exchange%20particles&f=false</a></p>
<p>What gives?</p>
|
<p>There are, essentially, three types of ensembles used in thermodynamics:</p>
<ul>
<li><strong>Microcanonical Ensemble:</strong> This is used to describe closed systems. The number of particles and the total energy are constant (since no interaction with the environment takes place). In such a system, the entropy will eventually be maximized.</li>
<li><strong>Canonical Ensemble:</strong> Now, your system is in contact with a big reservoir and allowed to exchange <em>energy</em> with the system. This means that your system will be in thermal equilibrium with the bath. </li>
<li><strong>Grand Canonical Ensemble</strong> Now, your system can exchange both energy and particles with the system. Hence, it will be in thermal equilibrium with the bath <em>and</em> in chemical equilibrium, i.e. the chemical potential for adding a particle to your system equals the chemical potential for adding a particle to the bath.</li>
</ul>
<p>If exchanging particles wasn't allowed in the grand canonical ensemble, it would become the canonical ensemble.</p>
| 250
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thermodynamics
|
Work for constant volume compression by adding moles of gas
|
https://physics.stackexchange.com/questions/9189/work-for-constant-volume-compression-by-adding-moles-of-gas
|
<p>I often see the expression $W = V \Delta P$ for the work of a constant volume compression where there are a fixed number of moles and the compression is caused by heating. Is this the work equation for a constant volume, isothermal process where the pressure is increased by adding moles of a gas?</p>
|
<p>In classical gas dynamics, in order to have work, you need a volume change $dV$. If volume $V$ is constant, no work is performed.</p>
<p>The traditional picture is to think of a piston closing a cylinder containing the gas. The gas inside the cylinder will be characterized by pressure $p$, temperature $T$ and volume $V$. To actually perform work, the piston needs to be moved, which implies a change of volume $dV=S\, dx$, where $S$ is the moving surface of the piston and $dx$ the distance it moved. Work is then performed by the force pushing the piston, related to pressure by $F=p\cdot S$, and during a slight motion $dx$ of the piston, work performed is
$dW=F\cdot dx=p\,S\,dx=p\, dV$. Adding work on small motions yields an integral, $W=\int p\, dV$. </p>
<p>If volume doesn't change, no work is performed. Energy may still be transfered into or out of the "cylinder" (whatever the gas' container is), but in a disorderly, non-directional fashion, that is as Heat, not Work.</p>
| 251
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thermodynamics
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free adiabatic expansion
|
https://physics.stackexchange.com/questions/9494/free-adiabatic-expansion
|
<p>My question is mainly an engineering question. Assume I have a turbine in which I adiabatically expand compressed air. The air cools down and does work to its surroundings, which is captured by the blades of the turbine and then is transformed into rotational energy/electricity. </p>
<p>When instead I let the compressed air flow through a nozzle into open space (no vacuum) and it expands adiabatically, it also cools down. But where is the internal energy of the air going? It does work to its surroundings, but where does this work end up? Does the surrounding heat up? Does it produce wind, i.e. kinetic energy in the surrounding air?</p>
<p>Feel free to replace adiabatic with polytropic. Whenever the gas cools down, I ask: Where does the energy end up, when I do not capture it with a turbine?</p>
|
<p>Adiabatic expansion in a nozzle results in internal energy being converted into kinetic energy. A generic steady-state energy balance is:</p>
<p>\begin{align}
\Delta (H + \frac{1}{2}{v^2} + zg) = Q + W
\end{align}</p>
| 252
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thermodynamics
|
A fan in a hot room at what point does it put in more energy that it dissipates
|
https://physics.stackexchange.com/questions/11428/a-fan-in-a-hot-room-at-what-point-does-it-put-in-more-energy-that-it-dissipates
|
<p>If a fan that is using 50 watts is moving 1 m³/min of air. Lets say the walls are the same temperature as the air so there’s no heat dissipation there. How do I know if the fan is putting in more energy in to the room than its dissipating?</p>
<p>If the walls are 10 deg cooler than the air will this be sufficient for the molecules to dissipate the heat in to them. Where’s the point of balance.</p>
|
<p>Assuming no heat transfer from a cooler outside to a warmer inside, a fan always adds more energy than it dissipates (because it dissipates no energy). However, by introducing a breeze, it accelerates evaporation off your skin, which in turn cools you off. (This is related to the wind chill factor meteorologists talk about.) How much cooler you will feel will be a function of how much you are sweating. It's important to note that pets, who don't sweat from the majority of their skin like we do, thus aren't cooled as much by a fan as we are.</p>
<p>Another method by which a fan can cool (as mentioned by Mitchell) is to move hot air away from a hot object, thus reducing the air's insulatory nature. This is why fans help cool off CPUs even though they don't sweat.</p>
| 253
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thermodynamics
|
Trying to understand a step in deriving Maxwell-Boltzman statistics
|
https://physics.stackexchange.com/questions/15129/trying-to-understand-a-step-in-deriving-maxwell-boltzman-statistics
|
<p>In the Wikipedia article on <a href="http://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_statistics" rel="noreferrer">Maxwell–Boltzmann statistics</a>, there is a point in the derivation that stumps me. When I get to where</p>
<p>$\displaystyle W=N!\prod\frac{g^{N_i}}{N_i!}$</p>
<p>is quoted as a count of the number of possible ways particles can be found in each state, I am told we want to pick the $N_i$ in such a way as to maximize $W$. Why is it important to maximize $W$?</p>
|
<p>That's the main assumption behind a lot of equilibrium statistical mechanics, i.e. the thermal equilibrium macrostate of a system is the one that can be realized in the largest possible amount of different microstates under some general restrictions on total energy and total amount of particles.</p>
<p>Or in other words, the thermal equilibrium state is assumed to be the one with the largest phase space volume. If you look at the total phase space of your system and start dividing it up into volumes with different macroparameters (i.e. different distributions of the particles over energy levels) you'll find out that there are small cells (for instance all particles in the highest energy level) and larger cells. The largest one is the one the system is most likely to be in most of the time and therefore it is natural to assume that it is that one that corresponds to thermal equilibrium. </p>
| 254
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thermodynamics
|
Is the cooling rate of a (very) cold object, sitting next to an AC higher or lower?
|
https://physics.stackexchange.com/questions/16076/is-the-cooling-rate-of-a-very-cold-object-sitting-next-to-an-ac-higher-or-low
|
<p>In more detail:
If i have two soda cans, both are cooled to exactly 4 degrees celsius,
And i put one in a 25 degrees room, and the other next to an AC vent set to 16 degrees.
After three minutes, which one should be colder than the other and why?</p>
<p>Edit: To clarify - if I have a cold soda can, should I place it near the AC vent or not (if I like my drink cold)? Which location will cause faster heating?</p>
|
<p>At the level you are asking the question it should be obvious that the closer to the cold output of the AC the can is sitting the cooler it will remain, since next to the vent the temperature will be the coldest and not yet mixed with the air in the room, a "refrigerator" set up.</p>
<p>The soda can when coming out of the refrigerator will be 4C but the air coming right out of the AC is much warmer, 16C, the close can will heat up to that in equilibrium. If the other can is at 25C it will reach an equilibrium at that temperature and will be warmer.The argument still holds for 3 minutes, the one at 25C will be incrementally warmer than the one at 16C.</p>
<p>If you are interested in the physics basis of this and real numbers, here is a chapter on <a href="http://canteach.candu.org/library/20043804.pdf" rel="nofollow">heat transfer.</a>: conduction, convection and radiation. Boundary conditions are necessary to get real numbers from differential equations.</p>
| 255
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thermodynamics
|
Thermodynamic System: Windmill
|
https://physics.stackexchange.com/questions/16976/thermodynamic-system-windmill
|
<p>There is a closed system that is just a windmill attached to a rotating shaft that can mechanically power something. I know the system produces work on the surroundings from the rotating shaft and this is positive work, but is there also a negative work input from the wind pushing the blades of the windmill?</p>
<p>Also, would there be any heat transfer in this scenario?</p>
|
<p>Let's start, as per the OP's comments, with the convention that when we take two systems A and B, and A is doing work on B, then: from the perspective of A, the work is positive; and from the perspective of B, the work is negative.</p>
<p>Now, from the perspective of the windmill. the work that the wind does is indeed negative: the wind is adding exergy to our windmill system. And most of that work will then transfer into positive work done by the windmill through the rotating shaft.</p>
<p>However, these transfers of exergy don't come for free: some heat is generated in the process, from friction in the system. As to exactly where the heat transfers are, will depend on the initial temperatures of the surroundings and the windmill, and the amount of heat generated by friction. But typically, the friction would raise the windmill's temperature above the temperature of the surroundings, so heat would get transferred from within the windmill system, to outside it.</p>
| 256
|
thermodynamics
|
Meaning of what C. P. Snow said about the laws of thermodynamics
|
https://physics.stackexchange.com/questions/17383/meaning-of-what-c-p-snow-said-about-the-laws-of-thermodynamics
|
<p>My native language is not English and I am having hard time understanding the meaning of the following statements which are to explain the laws of thermodynamics. </p>
<ol>
<li><p>The 1st law: you cannot win (meaning that I cannot make a profit?)</p></li>
<li><p>The 2nd law: you cannot break even (I looked up that expression but found many meanings)</p></li>
<li><p>The 3rd law: you cannot get out of the game (what game?)</p></li>
</ol>
|
<p>The general metaphor here is that you're gambling.</p>
<p>"You can't win." Yes, this means your profit can't be greater than zero. You go to your friend's house to play poker, and when you go home that night, you can't end up with more money (energy) than you started with.</p>
<p>"You can't break even." To "break even" means to go home after gambling with the same amount of money you had originally. Saying you can't break even means that you can't have zero profit. Entropy will increase, so the amount of energy available to do mechanical work must decrease.</p>
<p>"You can't get out of the game." The game is the same gambling game being implied in the first two metaphors. The application of the metaphor to the third law doesn't completely make sense to me; maybe it's just not an exact metaphor. Various sources online give interpretations like "because absolute zero is unattainable" or "there is no way to escape rules 1 and 2 because it is impossible to reach absolute zero." It makes sense to me that lowering an object's temperature to absolute zero -- which is impossible by the third law -- would in some sense mean that that object was "out of the game" thermodynamically, but it still wouldn't constitute a counterexample to the first or second laws.</p>
| 257
|
thermodynamics
|
how much heat i need to reach constant temperature
|
https://physics.stackexchange.com/questions/17686/how-much-heat-i-need-to-reach-constant-temperature
|
<p>Is there an equation I can use to calculate the heat required (as a function of time) to sustain water at some specif temperature?</p>
<p>I already know how much heat I need to reach the desired temparature using the formula Q=m*cp *(T2-T1). My question is how much heat do I need to keep 30,000lts of water at 70degress celcius for 30min once i have reached such temperature.</p>
<p>Any help would be greatly appreciated</p>
|
<p>If your 30,000 litre tank of water is not losing heat to the surroundings, then the power you need to apply to your tank is zero. While "zero heat loss" might or might not be realistic, it is important to understand this answer for this scenario.</p>
<p>If your 30,000 litre tank of water is losing heat to the surroundings (this could be by conduction through the tank walls, or airflow over an exposed surface (which to richen things up, might include some evaporative effects too)), then to hold the temperature of the tank constant you need to exactly replace the heat that is being lost.</p>
<p>It could be that what you really want to know is "how do I estimate the heat loss from a 30,000 litre tank". This is mostly of an engineering question, your best way to get a good answer might be to consult some engineering handbooks. The answer will depend on the size and shape of the tank, the conductivity of the tank's walls, whether or not the top is open or closed, whether the air surrounding the tank is still (so the heat transfer to the air is via natural convection only) or whether it is in motion (exposed to wind, fans, etc).</p>
| 258
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thermodynamics
|
In the Niagara Falls, which factors prevent rise of T H2O falling a certain height?
|
https://physics.stackexchange.com/questions/21395/in-the-niagara-falls-which-factors-prevent-rise-of-t-h2o-falling-a-certain-heig
|
<p>I need some ideas on a problem. </p>
<p>The first part says: Whats the posible rise in the temperature of the water falling 49.4 m in the Niagara Falls? That one was easy, with answer 0.112 Kelvin. ($\Delta T = \frac{g*h}{c_{H_2 O}}$)</p>
<p>The second part asks what factors tend to prevent that rise in temperature? Im thinking kinetic energy, preassure, air conductivity but im not sure. </p>
<p>Thanks in advance </p>
|
<p>The conservation of energy means that the potential energy liberated in falling must be present as heat. But there are precisely three ways for the water to lose that heat</p>
<ul>
<li>conduction/convection: by contact with air</li>
<li>radiatively: by emitting predominantly infrared light.</li>
<li>evaporation: by losing vapor into the air as the water is falling.</li>
</ul>
<p>From experience sweating, it should be apparent that in a fast air-flow environment, evaporative cooling is much more important than the other two effects. But this list is exhaustive--- there is no other place you can put the energy liberated by the fall.</p>
| 259
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thermodynamics
|
Boltzmann Brain Immortality
|
https://physics.stackexchange.com/questions/23013/boltzmann-brain-immortality
|
<p>This is a somewhat philosophical question. Given that even after person A's death, there's a non-zero chance that a brain fluctuates into existence with exactly the same structure and memories as A's brain, does that imply that A is immortal as given enough time, even if for each fluctuation, it survives for only a short while, A's brain will fluctuate into existence arbitrarily many times and this will enable A to "exist" for an arbitrarily long time.</p>
|
<p>Your question engages many of the same philosophical issues as those raised by so-called quantum immortality (http://en.wikipedia.org/wiki/Quantum_suicide_and_immortality). Discussions of quantum immortality are generally centered around the many worlds interpretation of quantum mechanics. The introduction of Boltzmann brain fluctuations into the discussion is an interesting variation. In any case, this is a question that lends itself more to the tools employed by philosophers than those employed by physicists (although this doesn't stop some physicists from devoting attention to it). </p>
| 260
|
thermodynamics
|
About heat rate and dimensions convention
|
https://physics.stackexchange.com/questions/28074/about-heat-rate-and-dimensions-convention
|
<p>Suppose I have a ceiling and its dimension are given in $Z \times C \times Y$, and a it's got a thermal conductivity $k$. Now, I want to know H, or the heat rate. It's given by $ H =\frac{kA(T_h -T_c)}{L}$. Where $T_h \ and \ T_c$ are the hot and cold temperature difference. How do I use the given dimensions for $A$ and $L$? Can I just assume $Z \times C$ is the area $A$ and $Y$ is $L$?</p>
|
<p>$L$ is thickness of the the layer and $A$ is area of the layer between two temperatures. The answer depends on what $Z$, $C$ and $Y$ stand for.</p>
| 261
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thermodynamics
|
How does the entropy of a system change when the amount of substance changes?
|
https://physics.stackexchange.com/questions/30039/how-does-the-entropy-of-a-system-change-when-the-amount-of-substance-changes
|
<p>I am thinking of a closed system consisting only of an ideal gas. Suddenly, the amount of substance of the gas changes. No other property of the system shall change.</p>
<p>How does this affect the entropy of the system? How can I calculate the change in entropy?</p>
|
<p>Entropy is an <a href="http://en.wikipedia.org/wiki/Intensive_and_extensive_properties" rel="nofollow">extensive</a> property. If you double the size of your system but keep the conditions within it the same then you will double the entropy. You can't do this without changing other properties of the system because lots of things, e.g. free energy, are also extensive.</p>
<p>If you keep the size of your system constant but pack in twice as much gas then the entropy will change for two reasons, firstly because you have more gas, but secondly because you've changed the pressure and/or temperature. You need to specify the state of your system after addition of the gas to work out the change in the entropy.</p>
| 262
|
thermodynamics
|
How does a lower elevation volume affect the temperature of the volume above it?
|
https://physics.stackexchange.com/questions/413865/how-does-a-lower-elevation-volume-affect-the-temperature-of-the-volume-above-it
|
<p>Assume you have a reasonably well thermally insulated box of air which you want to keep warm. It has at least one heat source, and it has at least one heat sink. One of the options available to you is to attach a second volume to the first, on the bottom of it, hoping to rely on the idea that the coldest air will collect in the lower volume that you do not care about.</p>
<p>Example:</p>
<pre><code>+---------------+
|Volume 1 | <-- at least one each of heat sink/source in here
| S |
+------------+ |
^ | |
heat | |V2| <-- Volume 2 added hoping to keep Volume 1 warmer
source(S)| | |
+--+
</code></pre>
<p><strong>How do we figure out what affect adding the second volume will have on the temperature of the first, upper volume?</strong></p>
<p><strong>For context</strong>, this question was actually raised initially from the topic of winter camping in cold areas. Someone had heard of advise to dig a hole straight down through the snow in the bottom of your igloo to give cold air a lower place to go that won't bother you, keeping the rest of the igloo slightly warmer. We are skeptical, but we are hoping to approach this logically instead of just dismissing the idea. We want to figure out how to estimate the effect that such a hole would really have.</p>
<p>So the heat source is human bodies resting on the floor of the upper volume, and the heat loss is due to various things including draft from door and ventilation holes and from loss through insulation (not very significant I don't think, as the snow actually insulates surprisingly well).</p>
| 263
|
|
thermodynamics
|
Heating effect of current
|
https://physics.stackexchange.com/questions/414968/heating-effect-of-current
|
<p>Why must the material of heating element of an electric heater habe high resistivity ? I mean, more resistivity, more resistance and hence less current and current is directly proportional to heating effect. And ive read that it is because if current is increased then resistance will decrease but heating effect is proportional to square of current and single term of resistance. So if high resistivity then very less current right? In that case heating effect will be less too. Decrease in current should decrease heat more than decrease in resistance does. So why are high resistivity materials used then? </p>
|
<p>Whatever current is flowing in the heating element will also have to flow in the wires leading to the heating element, including the wires in a transformer, which could be very long. </p>
<p>So, if we want to minimize heat losses in the wires, the resistance of the heating element has to be substantially higher than the resistance of the wires and therefore, for a reasonable geometry of the heating element, its material has to have high resistivity. </p>
| 264
|
thermodynamics
|
Understanding the meaning of thermodynamic work more clearly
|
https://physics.stackexchange.com/questions/419911/understanding-the-meaning-of-thermodynamic-work-more-clearly
|
<p>Consider a system having spherical symmetry filled with gas. If it expands or compress, there is some work done by system or surroundings, respectively, according to the equation $W=nRT\log(v_2/v_1)$. But the coordinates of centre of mass of system is same at the centre of sphere, so according to original definition of work ($W$ equal to force times displacement) it should be zero. I can't understand where is the mistake.</p>
|
<p>You need to understand the difference between boundary work on a closed system (expanding or compressing the sphere) vs work on or by an external force on the system as a whole (changing the velocity or position of the sphere). The former does not change the coordinates of the center of mass, but the latter does. It’s the difference between the change in internal energy of the system related to expansion or contraction of the boundary of the system, and the change in external energy of the system related to the change in velocity and/or position of the center of mass of the system with respect to an external frame of reference.</p>
<p>The long version of the first law for a closed system is:</p>
<p>Q – W = ΔE = ΔU + ΔKE + ΔPE</p>
<p>Where</p>
<p>ΔE = Total energy change of the system, which is the sum of change in internal and external energy of the system.</p>
<p>ΔKE = Change in kinetic energy of the system as a whole. This relates to a change in the velocity of the center of mass. By the work energy principle:</p>
<p>F x d = ΔKE</p>
<p>ΔPE = Change in potential energy of the system as a whole, such as a change in elevation of the center of mass (change in gravitational potential energy).</p>
<p>Q and ΔU are as always.</p>
<p>W now includes both boundary work and work done on or by the system a whole.</p>
| 265
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thermodynamics
|
Co2 cartridge questions
|
https://physics.stackexchange.com/questions/420278/co2-cartridge-questions
|
<p>When co2 is released from the cartridges. Is the gas get cold too or just the cartridge itself only ? And will be releasing it slowly will effect the temper drop ? Thanks you</p>
|
<p>It is an imprecise question. But then, once we know exactly what question to ask, we almost always have the answer already. So, let me try.</p>
<p>First of all, the $CO_2$ inside a cartridge is compressed to the point where it is a liquid at around room temperature. When gas is released, some of the liquid boils off.</p>
<p>It takes energy to convert from the liquid to the gaseous phase (heat of evaporation, or heat of vaporization). This is by the far the biggest effect. But watch out. The boiling off gas is created at the temperature of the liquid. However, boiling off the gas requires energy, which comes from the heat stored in the liquid - and the liquid will cool down.</p>
<p>As long as there is liquid $CO_2$ inside the cartridge the boiling off gas is created at the vapor pressure determined by the temperature of the liquid. As the gas expands adiabatically from around 70atm (near room temperature) to 1atm, it cools due to the Joule Thomson effect. However, this is a small effect. While it is being used in liquefying air (Linde method) it is not great for cooling a massive tank filled with a liquid, say water.</p>
<p>To answer the other question: The $CO_2$ liquid inside the cartridge is in close thermal contact with the cartridge wall and hence at the same temperature. If you drop the cartridge into the tank, the liquid $CO_2$ will equal the tank temperature after some time.</p>
<p>From here on we enter the realm of a poorly defined engineering problem. For instance, if the liquid in your tank is too hot, the cartridge may not be able to contain the $CO_2$ vapor pressure and start discharging uncontrollably.</p>
<p>If you put the cartridge into an ice bath outside your tank, you would start with $CO_2$ gas near 0°C, and you could send that through copper tubing through your tank.</p>
<p>Sources: <a href="https://en.wikipedia.org/wiki/Carbon_dioxide_(data_page)" rel="nofollow noreferrer">$CO_2$ thermodynamic data</a>,
<a href="https://www.nikhef.nl/~h73/kn1c/praktikum/phywe/LEP/Experim/3_2_06.pdf" rel="nofollow noreferrer">Joule Thomson Phyics Lab</a>, <a href="https://en.wikipedia.org/wiki/Hampson%E2%80%93Linde_cycle" rel="nofollow noreferrer">Hampson-Linde cycle</a></p>
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thermodynamics
|
A hot bar under tap water loses its heat by?
|
https://physics.stackexchange.com/questions/424157/a-hot-bar-under-tap-water-loses-its-heat-by
|
<p>In my book,it is given that hot bar placed under a running tap loses heat by conduction. My question is that why it can't be convection as the movement of water molecules is involved.</p>
|
<p>The cooling here obviously involves the movement of material, so, by definition, it is a form of convection and probably could be classified as a forced convection. </p>
<p>Of course, some thermal conduction also takes place, for instance, carrying heat from the bulk of the bar to its surface and from its surface to the passing water, but the dominant cooling mechanism here is due to the constant supply of cool water and therefore, it definitely involves convection. </p>
| 267
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thermodynamics
|
Obtain Magnetization from derivatives
|
https://physics.stackexchange.com/questions/432120/obtain-magnetization-from-derivatives
|
<p>I should determine the spontaneous magnetization of an ferromagnet below its critical temperature <span class="math-container">$T_{c}$</span> by only knowing the derivatives:</p>
<p><span class="math-container">$$
\left ( \frac{\partial M}{\partial H } \right )_{T}=\frac{a}{1-T/T_c}+3bH^{2}
$$</span>
where a and b are some real constants</p>
<p><span class="math-container">$$
\left ( \frac{\partial M}{\partial T } \right )_{H}=
\frac{1}{T_{c}}\frac{f(H)}{(1-T/T_c)^{2}}-
\frac{1}{2}\frac{M_{0}}{T_{c}}\frac{1}{(1-T/T_c)^{1/2}}
$$</span></p>
<p>with <span class="math-container">$M_{0}$</span>,<span class="math-container">$T_{c}$</span>,a and b are constants and f(H) is some function
with the property f(H=0)=0</p>
<p>a)determine f(H)
b)determine M(T,H)</p>
<p>Now I would start by writing down</p>
<p><span class="math-container">$$
dM = \left ( \frac{\partial M}{\partial H } \right )_{T}dH + \left ( \frac{\partial M}{\partial T } \right )_{H} dT
$$</span></p>
<p>This tells me how I would obtain the M in terms of the derivatives but I don't know how I would obtain the function f(H). Can anyone give me a hint how I would start the determination of f(H).</p>
<p><span class="math-container">$\mathbf{Edit}$</span>:
I tried now to do the integration what gives:
<span class="math-container">$$
M(T,H)=\frac{aH}{1-T/T_{c}}+bH^{3}+\frac{f(H)}{1-T/T_{c}}+M_{0}(1-T/T_{c})^{1/2}
$$</span></p>
<p>If I compute now again the derivative with respect to H I obtain:
<span class="math-container">$$
\left ( \frac{\partial M}{\partial H } \right )_{T}=\frac{a}{1-T/T_c}+3bH^{2}+\frac{1}{1-T/T_{c}}\frac{\partial f(H)}{\partial H}
$$</span></p>
<p>This is now not the same as in the exercise sheet. From here I don't know how to go on. Is this already wrong or is it somehow possible from this equation to determine f(H).
My first intuition was to say f(H) is just zero, because then the derivative with respect to H would be satisfied and so would be the f(H=0)=0 criterion</p>
|
<p>I was asking the same question in the mathematics forum and they figured out an answer for me:
<a href="https://math.stackexchange.com/questions/2940391/obtain-function-from-its-partial-derivatives">https://math.stackexchange.com/questions/2940391/obtain-function-from-its-partial-derivatives</a></p>
<p>For completeness I will also sketch this answer here.
Starting by doing the integration of the partial derivative with respect to H</p>
<p><span class="math-container">$$
M(H,T)=\frac{aH}{1-T/T_{c}}+bH^{3}+\phi(T)
$$</span></p>
<p>Now making the derivative with respect to T:</p>
<p><span class="math-container">$$
\frac{\partial M(T,H)}{\partial T}=\frac{1}{T_{c}}\frac{aH}{(1-T/T_{c})^{2}}+\phi'(T)
$$</span></p>
<p>By comparing the coefficients with the supplied derivative.
<span class="math-container">$f(H)=aH$</span> and <span class="math-container">$\phi'(T)=-\frac{1}{2}\frac{M_{0}}{T_{c}}\frac{1}{(1-T/T_{c})^{1/2}}$</span>.</p>
<p>The rest is then obtained by just inserting for f(H).</p>
| 268
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thermodynamics
|
Is the work done by friction considered as heat?
|
https://physics.stackexchange.com/questions/434405/is-the-work-done-by-friction-considered-as-heat
|
<p>In the first law of thermodynamics, We write the change in internal energy in terms of heat exchanged and work done. The answer to the above question doesn't matter if our final aim is to find the change in internal energy. However, When calculating the efficiency of an irreversible cycle, We always write difference of the heat taken in and heat rejected to be the work done by the system. In this, are we also including the work that is being done against friction?</p>
|
<p>Work done by friction is still work rather than heat. It is worth noting however that it is an <em>irreversible</em> process, not a reversible form of work.</p>
<p>In terms of efficiency, however, we are normally interested in "useful" work done by the system, or work done by the system on some other external system and so would not count work done against friction.</p>
| 269
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thermodynamics
|
Thermal expansion and thermodynamics
|
https://physics.stackexchange.com/questions/441481/thermal-expansion-and-thermodynamics
|
<p>Isn't thermal expansion and laws of thermodynamics contradictory?
Gas expands when heated, but after expansion it cools down.</p>
|
<p>One of the first things to learn in thermodynamics is the importance of explicitly stating the conditions of a transformation. Speaking of thermal expansion and cooling, without specifying how those processes happen allows almost any answer to be correct or false.</p>
<p>For example, a heated gas expands in a container with a movable wall, <strong>if</strong> the heating makes its pressure higher than the external pressure. The cooling you mention can be due to the expanded gas's energy loss if the environment is at a lower temperature. Still, it is also happening when you make a real gas expanding in a surrounding environment at a lower pressure (like the case of compressed gas in an aerosol can when one presses the nozzle).</p>
<p>It is the variety of possible conditions that may induce the idea of contradiction. Actually, the whole manifold of such different behaviors can be nicely explained by thermodynamics.</p>
| 270
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thermodynamics
|
Why isn't $dH=Q+\sum_i\mu_idN_i$ (constant pressure)
|
https://physics.stackexchange.com/questions/446187/why-isnt-dh-q-sum-i-mu-idn-i-constant-pressure
|
<p>So my book claims that under constant-pressure conditions, the change in enthalpy is given by <span class="math-container">$dH=Q$</span>. However, if we look at the thermodynamic identity and first law of thermodynamics, we see that
<span class="math-container">$$
dU=TdS-PdV+\sum_i\mu_idN_i=Q+W.
$$</span>
This would mean that
<span class="math-container">$$
dH= dU+PdV=TdS+\sum_i\mu_idN_i=Q+W+PdV=Q+\sum_i\mu_idN_i.
$$</span>
Since heat is defined as the spontaneous flow of energy due to a temperature difference, I would say that we can't include the term <span class="math-container">$\sum_i\mu_idN_i$</span> to <span class="math-container">$Q$</span> (because it represent the change in energy by adding particles), but then <span class="math-container">$dH$</span> isn't equal to <span class="math-container">$Q$</span>... Besides, my book calls <span class="math-container">$\sum_i\mu_idN_i$</span> "chemical work", which seems to suggest that it should belongn to <span class="math-container">$W$</span> and not <span class="math-container">$Q$</span>.</p>
<p>Could someone help me out of my confusion?</p>
<p><strong>EDIT</strong></p>
<p>I wasn't clear about this at first, but question kind of revolves around chemical reactions. So we're talking about a closed system, where the chemical work is nonzero.</p>
|
<p>The chemical potential <span class="math-container">$d\mu_i$</span> is defined as the partial molar free energy: <span class="math-container">$$\mu_i=\left(\frac{\partial G}{\partial N_i}\right)_{T,P}$$</span> But, as a result of this and the other basic relationships, it is also given by:
<span class="math-container">$$\mu_i=\left(\frac{\partial U}{\partial N_i}\right)_{S,V}=\left(\frac{\partial H}{\partial N_i}\right)_{S,P}$$</span></p>
<p>So, <span class="math-container">$$dH=TdS+VdP+\sum{\mu_idN_i}$$</span> and, at constant pressure,<span class="math-container">$$dH=dU+PdV=TdS+\sum{\mu_idN_i}$$</span></p>
<p>But, dQ is not equal to TdS unless the process is reversible.</p>
<p>Also, please don't mix differentials up with finite changes. A finite change should be represented by a <span class="math-container">$\Delta$</span>. And Q is not equal to <span class="math-container">$T\Delta S$</span> unless the temperature is constant (and, again, unless the process is reversible).</p>
<p>Also, <span class="math-container">$$\Delta H=Q$$</span> only applies to a closed system (i.e., no mass entering or leaving). Obviously, if mass is entering or leaving, it does so with its own enthalpy (and other properties), and this changes the enthalpy of the system.</p>
| 271
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thermodynamics
|
Caloric equation of state for the Van der Waals gas
|
https://physics.stackexchange.com/questions/453594/caloric-equation-of-state-for-the-van-der-waals-gas
|
<p>I'm currently trying to reproduce a specific derivation of the caloric equation of state for the Van der Waals gas, which I saw a couple of months ago in a Thermodynamics lecture. I'm well aware of the fact that there are already multiple derivations on this site and online (see <a href="https://physics.stackexchange.com/questions/46737/internal-energy-according-to-the-van-der-waals-equation">[1]</a>, <a href="https://physics.stackexchange.com/questions/100974/finding-internal-energy-using-equation-of-state">[2]</a>, <a href="https://physics.stackexchange.com/questions/46737/internal-energy-according-to-the-van-der-waals-equation">[3]</a>, <a href="https://www.physicsforums.com/threads/derive-the-energy-equation-for-a-van-der-waal-gas.744397/" rel="nofollow noreferrer">[4]</a>, <a href="https://www.thestudentroom.co.uk/showthread.php?t=4043485" rel="nofollow noreferrer">[5]</a>), but none of them are what I'm looking for. </p>
<p>The things that I remember from the derivation:</p>
<ol>
<li>The inner energy was a function of <span class="math-container">$T$</span> and <span class="math-container">$v$</span>.</li>
<li>We used the fact that <span class="math-container">$s$</span> is a state variable.</li>
<li>After using the first law we expanded all differentials in some variables.</li>
<li>With this we obtained <span class="math-container">$u(T)=f(T)-a/v$</span> and then proceeded to find <span class="math-container">$f(T)$</span> by using the fact that <span class="math-container">$C_V = (\delta Q/\partial T)_V = (\partial u/\partial T)_V$</span>.</li>
</ol>
<hr>
<p><strong>What I've tried so far:</strong> Using the first law we obtain <span class="math-container">$Tds=du+pdv$</span>. We can express <span class="math-container">$du$</span> and <span class="math-container">$dv$</span> in terms of <span class="math-container">$T,p$</span>, which leads to the equation
<span class="math-container">$$ds = \frac{1}{T}\left(\frac{\partial u}{\partial p}\Bigg|_{T} + p \frac{\partial v}{\partial p}\Bigg|_{T}\right)dp+\frac{1}{T}\left(\frac{\partial u}{\partial T}\Bigg|_{p}+ p\frac{\partial v}{\partial T}\Bigg|_{p}\right)dT.$$</span>
Using the fact that <span class="math-container">$s$</span> is a state variable we get the relation
<span class="math-container">$$-\frac{\partial v}{\partial p}\Bigg|_{T} = \frac{1}{T}\left(\frac{\partial u}{\partial T}\Bigg|_{p}+p \frac{\partial v}{\partial T}\Bigg|_{p}\right).$$</span>
This simplifies the first equation a bit but isn't really helpful since it's impossible to acutally take the derivative of <span class="math-container">$v$</span> with respect to any variable due to the form of the Van der Waals equation. </p>
<p>Since I couldn't figure anything out with this I tried the following as another approach:
<span class="math-container">$$\begin{align} ds &= \frac{1}{T} \left(\frac{\partial u}{\partial v}\Bigg|_{T}dv + \frac{\partial u}{\partial T}\Bigg|_{v}dT\right) + \frac{p}{T}dv \\
& = C_V \frac{dT}{T}+ \frac{1}{T}\left(\frac{\partial u}{\partial v}\Bigg|_{T}+p \right)dv.\end{align}$$</span>
This approach seems to similar to the one from <a href="https://physics.stackexchange.com/questions/46737/internal-energy-according-to-the-van-der-waals-equation">[1]</a>, but I'm 100% sure that we didn't use the Helmholtz equation as suggested by @juanrga. </p>
<hr>
<p><strong>TL;DR</strong> Could someone help me derive <span class="math-container">$u(T)=C_VT-a/v$</span> (caloric equaiton of state) for the Van der Waals gas without using any fancy equations, but rather fundamentals like the fact that <span class="math-container">$s$</span> is a state variable, etc. </p>
|
<p>I am really puzzled by the lack of a clear answer to this question, even after reading the references you cited.</p>
<p>The real answer (your ref. [5] goes quite near to it, but it misses the point as well) is simply that <strong>it is not possible to obtain it if the only information is the van der Waals equation of state!</strong> This is quite well explained in Callen's book on thermodynamics. A strong indication comes from the fact that, even in the perfect gas, the equation of state alone does not distinguish among mono- di- or poly-atomic perfect gases.</p>
<p>From the formal point of view, the root of the problem is trivially: to have <span class="math-container">$U(V,T)$</span> or any equivalent information, one has to integrate the pressure. Knowing <span class="math-container">$P(T,V,N)$</span>, we know the <em>partial</em> derivative of the Helmholtz free energy <span class="math-container">$F(T,V,N)$</span> with respect to <span class="math-container">$V$</span> <em>at fixed <span class="math-container">$T,N$</span></em>. The integration of
<span class="math-container">$$
P = - \left. \frac{\partial{F}}{\partial{V}} \right|_{T,N}
$$</span>
can provide <span class="math-container">$F$</span> only within an <strong>arbitrary function</strong> of <span class="math-container">$N$</span> and <span class="math-container">$T$</span>. Which is equivalent to know almost nothing about the thermal equation of state.</p>
<p>Also, arguments based on assuming that the limiting behavior at large volumes is the same as a perfect gas are flawed. Perfect gas is the limit of real gases at large volume (low pressure) <em>and</em> high temperature. Van der Waals equation of state is routinely used for describing fluids even close to their triple point. Extending so much the validity of the asymptotic regime is equivalent to <em>postulate</em> the thermal equation of state, and it is not proof.</p>
<p>The real answer is that knowledge of the equation of state <span class="math-container">$P=P(T,V,N)$</span> alone does not allow to obtain the caloric equation of state. Some additional information must be added, for example, information about the constant volume specific heat as a function of (T,V,N). In any case, it is an independent additional information.</p>
| 272
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thermodynamics
|
Will the particles from 2 objects in thermal equilibirum still collide?
|
https://physics.stackexchange.com/questions/453691/will-the-particles-from-2-objects-in-thermal-equilibirum-still-collide
|
<p>Suppose that we have 2 objects and both of them reached thermal equilibrium. Do the particles of the two objects still collide with each other? If so, do any of the collisions result in the transfer of energy between the two objects? Explain.</p>
|
<p>Microscopically the particles of the two bodies are of course in motion. Motions do not cease just because the two bodies have the same temperature, this just means that they happen with the same averaged energy.</p>
<p>Even if each individual body is first of all in internal equilibrium itself, the kinetic energy of its particles obey to a statistical distribution, some have more some less. As such there will be plenty of collisions that transfer energy between particles within the same body but without no net change.</p>
<p>The same happens at the boundary between the two bodies. Plenty of collisions transfer energy from, say, A to B, while others do the opposite.</p>
<p>Thus, yes there are collisions but not overall energy exchange. </p>
<p>Related Q&As here <a href="https://physics.stackexchange.com/questions/453680/does-heat-transfer-actually-from-high-temperature-to-low-temperature">Does heat transfer actually from high temperature to low temperature?</a></p>
| 273
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thermodynamics
|
How do all chemical reactions result in a product whose potential is less than the reactants if all reactions are not exothermic?
|
https://physics.stackexchange.com/questions/455551/how-do-all-chemical-reactions-result-in-a-product-whose-potential-is-less-than-t
|
<p>We are told by our chemistry teacher that all chemical reactions take place as reactants (that have bigger potential) change into products (that have smaller potential) hence unstable substances change into more stable ones. But doesn't this mean "all reactions are exothermic"? I am in need of a clarification, please.</p>
|
<p>"Potential," as your teacher is using it, is not exactly the same as energy. (S)he is likely talking about the Gibbs energy <span class="math-container">$G$</span>, which is a combination property</p>
<p><span class="math-container">$$G = H - TS$$</span></p>
<p>where <span class="math-container">$H$</span> is the enthalpy. If we think about the change in <span class="math-container">$G$</span>:</p>
<p><span class="math-container">$$ \Delta G = \Delta H - \Delta (T S) $$</span></p>
<p>One way of reducing <span class="math-container">$G$</span> is to reduce <span class="math-container">$H$</span> (which chemists usually think of as "exothermic"), but it is also possible to increase <span class="math-container">$H$</span> (which chemists usually think of as "endothermic") provided that you simultaneously increase the product <span class="math-container">$T S$</span> by more.</p>
<p>When labelling a reaction as endothermic/exothermic, chemists usually think about <span class="math-container">$\Delta H$</span>, but when labelling a substance as stable or unstable, they usually think <span class="math-container">$G$</span>.</p>
| 274
|
thermodynamics
|
Why would the gas do work when the piston is pulled by external force
|
https://physics.stackexchange.com/questions/459600/why-would-the-gas-do-work-when-the-piston-is-pulled-by-external-force
|
<p>Taking an adiabatic process as example , from the first law of thermodynamics <span class="math-container">$$\Delta U=-W... (1)$$</span> .Now if I pull up the piston applying force from my side , the work is done by me on the system so the sign is negative, so from (1) , <span class="math-container">$\Delta U$</span> is positive , but I know that temperature should decrease and so should internal energy , then is it that I am taking the wrong signs <br> moreover , why would the gas lose its internal energy anyway , after all , its the external force applied by me which is moving the piston , so why should the gas inside do any work and lose temperature ?</p>
|
<p>Imagine experiencing in a vacuum to avoid the issue of atmospheric pressure outside the piston. A wedge prevents the piston from mounting and is removed at the beginning of the experiment.</p>
<p>If you pull the piston very quickly, faster than the speed of sound, there will be no resistance: the outside work will be zero. It is a Joule expansion, without work and <span class="math-container">$\Delta U=0$</span>.</p>
<p>If you let the piston go up slowly, it is the gas that pushes the piston and you must exert a force directed downwards which prevents it from accelerating. In this case <span class="math-container">$-W<0$</span>, and therefore <span class="math-container">$\Delta U<0$</span> because the gas provides work.</p>
<p>Sorry for my english.</p>
| 275
|
thermodynamics
|
How to know the dependances of thermodynamic variables
|
https://physics.stackexchange.com/questions/463100/how-to-know-the-dependances-of-thermodynamic-variables
|
<p>In the proof of :</p>
<p><span class="math-container">$$dU=C_v dT - \left[T \left( \frac{\partial P}{\partial T} \right)_V - P\right] dV$$</span></p>
<p>on wikipedia <a href="https://en.wikipedia.org/wiki/Internal_energy" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Internal_energy</a>, they uses the fact that :</p>
<p><span class="math-container">$$dS=\left( \frac{\partial S}{\partial T} \right)_V dT + \left( \frac{\partial S}{\partial V} \right)_T dV $$</span></p>
<p>Then they replace the expression of <span class="math-container">$dS$</span> in the thermodynamic relation and they find out the good relationship.</p>
<p><strong>However</strong>, the thing that disturbs me is the dependance of the state function on the thermodynamic variables.</p>
<p>I thought that a state function must depend on <strong>extensive</strong> thermodynamic variables. When we write the expression of <span class="math-container">$dS$</span> here we have a <span class="math-container">$dT$</span> appearing on the rhs. This is what disturbs me.</p>
<p>So in all generality, how can we know that the expression of <span class="math-container">$dS$</span> in terms of its variable is the good one ?</p>
<p>Why can't we write for example :</p>
<p><span class="math-container">$$dS=\left( \frac{\partial S}{\partial T} \right)_{V,P} dT + \left( \frac{\partial S}{\partial T} \right)_{V,T} dP + \left( \frac{\partial S}{\partial V} \right)_{T,P} dV $$</span></p>
<hr>
<p><strong>My guess about it that I would like to check :</strong></p>
<p>I guess there is an implicit assumption that we know the thermodynamic potentials in our case (gas in a closed system) only depends on two thermodynamic variables. </p>
<p><strong>Then, first thing : we must only have two terms on the rhs of <span class="math-container">$dS$</span>.</strong></p>
<p>In principle, I could decide to express <span class="math-container">$S$</span> in function of the two intensives variables : <span class="math-container">$(U,V)$</span>. But I can always write <span class="math-container">$dU=\left(\frac{\partial U}{\partial T} \right)_V dT + \left(\frac{\partial U}{\partial V} \right)_T dV$</span></p>
<p><strong>Then, I can decide to express <span class="math-container">$S(T,V)$</span> instead of <span class="math-container">$S(U,V)$</span></strong> </p>
<p>Is it correct ?</p>
<p>But then, how do we know that the two independant variables will be <span class="math-container">$(T,V)$</span> and not <span class="math-container">$(P,V)$</span> for example ? How to know which two independant variable I can chose ? I guess it is linked to the fact I must consider non conjugated variables but I don't really see why.</p>
| 276
|
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thermodynamics
|
Heat transfer and steady state
|
https://physics.stackexchange.com/questions/477612/heat-transfer-and-steady-state
|
<p>Consider a metal rod between two bodies of different temperature.How do you explain the occurrence of steady-state when heat Is transferring by conduction through the rod. I mean why should there exist a point called steady-state in the first place, meaning why the temperature of the rod at a point become static in steady state?</p>
|
<p>The following is a very crude description of what is happening, but it captures the basic physical mechanisms in play. Suppose that the entire rod is initially at the temperature <span class="math-container">$T_c$</span> and at time t = 0, the left end of the rod x = 0 is suddenly raised to temperature <span class="math-container">$T_h$</span> (and held at that temperature forever afterward) while the right end of the rod x = L continues to be held at temperature <span class="math-container">$T_c$</span>.</p>
<p>Heat will begin to propagate into the rod (by conduction) from the left end, and, as time progresses, the "effective" depth of propagation <span class="math-container">$\delta(t)$</span> of the disturbed region will grow with time, until the depth of penetration encompasses the entire rod. A crude approximation to the temperature profile within the rod at any time can be taken as: <span class="math-container">$$T=T_h-(T_h-T_c)\frac{x}{\delta(t)}\tag{for x < delta}$$</span>and<span class="math-container">$$T=T_c\tag{for x > delta}$$</span>The total amount of heat that has entered the rod at any time is equal to <span class="math-container">$$\int_0^L{\rho C A(T-T_c)}dx$$</span> where <span class="math-container">$\rho$</span> is the rod density, C is the specific heat capacity, and A is the cross sectional area. For the assumed temperature profile, this becomes <span class="math-container">$$\rho C A (T_h-T_c)\frac{\delta(t)}{2}$$</span>Also, from the assumed temperature profile, the rate of heat flow into the rod at the left end is <span class="math-container">$$kA\frac{(T_h-T_c)}{\delta(t)}$$</span> and the rate of heat flow out of the rod at the right end is zero, where k is the thermal conductivity. So, a time-dependent heat balance on the rod gives:<span class="math-container">$$\frac{d}{dt}\left(\rho C A (T_h-T_c)\frac{\delta(t)}{2}\right)=kA\frac{(T_h-T_c)}{\delta(t)}$$</span>or, equivalently,<span class="math-container">$$\delta\frac{d\delta}{dt}=2\frac{k}{\rho C}$$</span>The solution to this differential equation is <span class="math-container">$$\delta(t)=2\sqrt{\frac{k}{\rho C}t}$$</span>So this equation tells us the effective distance of penetration of the thermal boundary layer into the rod as a function of time. Steady state is approximately attained when <span class="math-container">$\delta = L$</span>, or, equivalently, when <span class="math-container">$$t=\frac{\rho C}{k}\frac{L^2}{4}$$</span>
As I said, this is a very crude approximation to the solution to the transient heat conduction equation, but it includes all the important physical mechanisms.</p>
| 277
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thermodynamics
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Joule-Thomson effect: why does a gas cool if it's below the inversion temperature?
|
https://physics.stackexchange.com/questions/483132/joule-thomson-effect-why-does-a-gas-cool-if-its-below-the-inversion-temperatur
|
<p>The Joule-Thomson coefficient is given by
<span class="math-container">$$\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_{H} = \frac{V}{C_{P}}(\beta T - 1),$$</span>
where <span class="math-container">$\beta$</span> is the coefficient of thermal expansion. If the inversion temperature is defined by <span class="math-container">$T_{inv} = \frac{1}{\beta}$</span>, then why is <span class="math-container">$\mu_{JT} > 0$</span> if <span class="math-container">$T < T_{inv}$</span>, as stated by Wikipedia <a href="https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect" rel="nofollow noreferrer">Joule-Thomson effect</a>? I really don't get this, this should be basic math ? This results in not comprehending why a gas cools if it's below inversion temperature and vice versa.</p>
|
<p>If you look up the coefficient of thermal expansion of air, you will find that it decreases with increasing absolute temperature. In fact, it decreases rapidly enough that even the product <span class="math-container">$\beta T$</span> decreases with increasing temperature. For air at room temperature and 1 bar, for example, the product is about 1.01 while, at 200 C, the product is about 0.99.</p>
<p>In terms of the compressibility factor z, the product of <span class="math-container">$\beta$</span> and T is given by: <span class="math-container">$$\beta T=1+\left(\frac{\partial \ln{z}}{\partial \ln{T}}\right)_P$$</span>So <span class="math-container">$$(\beta T-1)=\left(\frac{\partial \ln{z}}{\partial \ln{T}}\right)_P$$</span></p>
<p>Get yourself a plot of z vs reduced temperature and reduced pressure and note how the right hand side of this equation decreases with increasing reduced temperature. It is positive at low reduced temperatures, and reaches a value of zero at a reduced temperature of about 4.</p>
| 278
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thermodynamics
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Change of pressure in Isothermal process
|
https://physics.stackexchange.com/questions/484915/change-of-pressure-in-isothermal-process
|
<p>In isothermal process if the internal energy is not increasing then how does the pressure change?</p>
<p>I come to the question because in isobaric process the pressure remains unchanged because the change of volume and temperature . But in isothermal process the the temperature is not changed but pressure is changing.</p>
|
<p><strong><em>In isothermal process if the internal energy is not increasing then how does the pressure change?</em></strong></p>
<p>It changes because in an isothermal process the product of pressure and volume is constant. For an ideal gas when heat is added to the system the system does an equal amount of work increasing in volume while decreasing in pressure. For an ideal gas where <span class="math-container">$PV=nRT$</span> this means temperature is a constant. Since the change in internal energy of an ideal gas depends only on temperature change, the change in internal energy is therefore zero. From the first law </p>
<p><span class="math-container">$$\Delta U=Q-W$$</span></p>
<p>Since <span class="math-container">$\Delta U=0$</span>, <span class="math-container">$Q=W$</span>. For a reversible expansion the heat added to the system exactly equals the work done by the system, causing the volume to increase and pressure to decrease such that the product remains constant. For a compression the work done on the system exactly equals the heat leaving the system, causing the volume to decrease and pressure to increases so that the product is constant. </p>
<p><strong><em>I come to the question because in isobaric process the pressure remains unchanged because the change of volume and temperature . But in isothermal process the the temperature is not changed but pressure is changing.</em></strong></p>
<p>In an isobaric process the heat that is added to the system does work expanding the system but it also increases the internal energy of the system. It is because some of the heat goes into increasing the internal energy that the pressure can remain constant while volume and temperature increases.</p>
<p>Hope this helps.</p>
| 279
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thermodynamics
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Can a non-quasi-static process have an equation of the path which it has followed?
|
https://physics.stackexchange.com/questions/485814/can-a-non-quasi-static-process-have-an-equation-of-the-path-which-it-has-followe
|
<p>Recently I saw a question where there was a specific p-v relation for a process, but it was stated that it's not a quasi-static process. Hence, we can not calculate the work done by integration of pdv work. </p>
|
<p>You can can calculate the work done by integrating <span class="math-container">$pdv$</span>, but the pressure is the external pressure, not the gas pressure. For a non quasistatic process the gas is not in equilibrium internally as pressure gradients exist. That makes the internal pressure undefined.</p>
<p>This will respond to your follow up question:</p>
<p><strong><em>But, how can it have a function of its path?</em></strong></p>
<p>For a non quasi-static process the function only represents the external applied pressure and the volume of the gas. All other properties of the gas are undefined during the process because the gas is not in internal equilibrium. Work is always <span class="math-container">$pdv$</span> where the pressure is the external pressure. Only if the process is carried out quasi statically is the internal gas pressure equal to the external pressure at each point of the process. </p>
<p>For example, for an ideal gas at equilibrium we have <span class="math-container">$PV=nRT$</span>. Since the gas is not in internal equilibrium during a non quasi-static process, this equation does not apply at each point of the process with the exception that <span class="math-container">$V$</span> in the function does represent the volume of the gas but temperature and pressure are not in internal equilibrium. In other words, the points on the process curve only represent equilibrium states of the gas if the process is carried out quasi statically. You can apply the ideal gas equation at the initial equilibrium state before the process starts and at the final state when the process stops and the gas is allowed to reach internal equilibrium, but not at the points in between. </p>
<p>Hope this helps.</p>
| 280
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thermodynamics
|
How can heat be a inexact differntial, in thermodynamics?
|
https://physics.stackexchange.com/questions/489305/how-can-heat-be-a-inexact-differntial-in-thermodynamics
|
<p>I came across this definition of the first law of thermodynamics in <a href="https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Differential_Forms_of_Fundamental_Equations" rel="nofollow noreferrer">Differential Forms of Fundamental Equations</a>:</p>
<p><span class="math-container">$$dU={\bar d}q+{\bar d}w,$$</span></p>
<p>where it states that the symbol <span class="math-container">$\bar{d}$</span> is an inexact differential. But I don't understand how heat is an inexact differential. </p>
<p>I mean work is because, work is dependent on its path, so it has a path function, but I never come across a path function for heat. Is there a path function for heat?</p>
|
<p>The amount of heat supplied during a given change of thermodynamic state depends on the path of the change.</p>
<p>For example, consider an ideal gas moving from <span class="math-container">$(p_1,V_1)$</span> to <span class="math-container">$(p_2,V_2)$</span>. There are an infinite number of routes it could follow. Let's consider two: </p>
<p><span class="math-container">$$
\mbox{either }\; (p_1,V_1) \rightarrow (p_2,V_1) \rightarrow (p_2,V_2)\\
\mbox{or }\;\;\;\;\;\; (p_1,V_1) \rightarrow (p_1,V_2) \rightarrow (p_2,V_2)
$$</span>
where the change indicated by each arrow falls on a straight line on a <span class="math-container">$p,V$</span> diagram.
The internal energy <span class="math-container">$U$</span> changes by the same amount for these two routes, since <span class="math-container">$U$</span> is a function of state. But the work done in these two routes is different: in the first case it is <span class="math-container">$p_2(V_1-V_2)$</span> but in the second case it is <span class="math-container">$p_1 (V_1-V_2)$</span>. It follows that the heat supplied differs also.</p>
<p>Now let's look a little more generally. For the case of irreversible heat transfer one can write
<span class="math-container">$$
\bar{d}Q_{\rm rev} = T\,dS
$$</span>
where <span class="math-container">$T$</span> and <span class="math-container">$S$</span> are functions of state. But there is no single-valued function whose differential is <span class="math-container">$T\,dS$</span>. Proof: suppose that there were such a function and call it <span class="math-container">$f$</span>. Then we have
<span class="math-container">$$
df = T\,dS \\
\Rightarrow \;\;
\left.\frac{\partial f}{\partial S}\right|_T = T, \;\;
\left.\frac{\partial f}{\partial T}\right|_S = 0 \\
\Rightarrow \;\;
\frac{\partial^2 f}{\partial T\,\partial S} = 1, \;\;
\frac{\partial^2 f}{\partial S\,\partial T} = 0.
$$</span>
Hence
<span class="math-container">$$
\frac{\partial^2 f}{\partial T\,\partial S}
\ne \frac{\partial^2 f}{\partial S\,\partial T}
$$</span>
But this is impossible for a well-behaved single-valued function. Hence we have that <span class="math-container">$T\,dS$</span> is not a proper differential. QED</p>
| 281
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thermodynamics
|
Is electromagnetic radiation released because of a difference in temperature or because instability of electrons inside matter?
|
https://physics.stackexchange.com/questions/489330/is-electromagnetic-radiation-released-because-of-a-difference-in-temperature-or
|
<p>A thermodynamic question: is electromagnetic radiation released because of a difference in temperature or because instability of electrons inside matter (atoms and molecules)?</p>
<p>The thermodynamic law states that energy goes from higher temperature to lower temperature in my study book, but I think that it's not the case for radiation. Could someone elaborate?</p>
|
<p>Temperature is just a measure of how fast the atoms/molecules of an object "jiggle". The concept of temperature makes sense only for large number of atoms/molecules. Because atoms are electrically charged, and moving charges emit electromagnetic radiation, it means that all objects with <span class="math-container">$T>0$</span> (<span class="math-container">$T$</span> is absolute temperature) will emit thermal radiation. The higher the temperature is, the higher the average kinetic energy of the atoms will be - thus the faster they will "jiggle". This is called <strong>thermal radiation</strong>.</p>
<p>Electromagnetic radiation can also be released when electrons make a transition from one higher energy state to a lower one. But this release of radiation happens for different reasons. The energy levels in atoms are quantized, meaning that the atom can occupy only particular energy levels: <span class="math-container">$E_{1}, E_{2}, E_{3},\dots E_{n}$</span>. For example, when an electron from an atom makes a transition from state <span class="math-container">$2$</span> to state <span class="math-container">$1$</span>, there will be a photon emitted in the transition with an energy of:
<span class="math-container">$$\hbar\omega=E_{2}-E_{1}$$</span></p>
<p>Now going back to thermal radiation and thermodynamics laws. Heat is just the energy transfer/exchange between thermodynamic systems. Heat transfer goes on average from hot object to the cold object because the atoms of the hot object have higher kinetic energy than the atoms of the cold one. </p>
<p><a href="https://i.sstatic.net/7IR1h.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7IR1h.gif" alt="enter image description here"></a></p>
<p>Just to make a simplistic analogy: you can view this process just like (billiard) balls colliding. When two balls collide, the one with higher kinetic energy will lose energy by transfering to the one with lower kinetic energy.
We can further imagine a frictionless billiard table with <span class="math-container">$N$</span> balls of different kinetic energies that are colliding with each other and with the table margins. In the end, given enough time, what will happen is that the energy will be equally distributed on the whole table.</p>
| 282
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thermodynamics
|
Heat transfer through window
|
https://physics.stackexchange.com/questions/493287/heat-transfer-through-window
|
<p>I was wondering what is the ideal number of glass layers in a windows assuming outside T2 is greater than inside T1 and we want T1 to remain as low as possible (ie. keep the room cool). Now heat capacity (C) and thermal conductivity (λ) of air between glass layers is lower than that of glass but having no glass is essentially having window open which obviously isn't correct solution. On the other hand stack of glass cannot be ideal as well. So there is some middle ground. Maybe theoretically there should be infinite number of as thin as possible layers of glass separated with air?</p>
<p>Thank you. </p>
| 283
|
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thermodynamics
|
Thermodynamic planks temperature
|
https://physics.stackexchange.com/questions/501196/thermodynamic-planks-temperature
|
<p>There is a limit of temperature that is about <span class="math-container">$0$</span> K = <span class="math-container">$-273.15$</span> <span class="math-container">$^\circ$</span>C. Is there any maximum limit?</p>
|
<p>Nope. Not really. (I need 30 characters) </p>
| 284
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thermodynamics
|
Queries about the carnot cycle
|
https://physics.stackexchange.com/questions/507397/queries-about-the-carnot-cycle
|
<p>So I was studying the carnot cycle and was taught it as the following sequence.</p>
<p>Imagine a cylinder filled with some ideal gas at <span class="math-container">$T_1$</span>. It has some rocks on top of it and the system is in equilibrium.
If you place it in a reservoir at temperature <span class="math-container">$T_1$</span>, and remove the small rocks one by one, you can treat this as an isothermal expansion. </p>
<p>Then follows an adiabatic expansion, by removing more small rocks while keeping the system thermally isolated.</p>
<p>After which point the system will reach a lower temperature <span class="math-container">$T_2$</span>. Placing it again in a reservoir at <span class="math-container">$T_2$</span> and adding rocks back will compress the gas isothermally. </p>
<p>The final part is an adiabatic compression back to the initial point in the phase diagram, done by adding more rocks while the system is kept in thermal isolation.</p>
<p>This is my understanding of a carnot cycle.</p>
<p>My question is that, how is this system doing any work? I understand how to calculate the work done on a phase diagram. The main concern is that isnt the work done by us in removing and adding the rocks balanced by the work done by the piston in moving up and down? Isn't the overall work done still 0?</p>
<p>Secondly what is the significance of the isothermal adiabatic expansion / contraction sequence. My previous impression was that any loop in a phase diagram was reversible if the process was done slowly. Is the efficiency of reversible processes not the same? What is it about the carnot cycles sequence that gives it the greatest efficiency?</p>
<p>Not sure if these questions are coherent but I am still not clear on why the carnot engine seems to be such a big deal.</p>
<hr>
|
<blockquote>
<p>The main concern is that isnt the work done by us in removing and
adding the rocks balanced by the work done by the piston in moving up
and down? </p>
</blockquote>
<p>Each rock is slid off the piston horizontally during the isothermal and adiabatic expansions, and slid back onto the piston horizontally, during the isothermal and adiabatic compressions. No work is associated with the horizontal movement of the rocks against or with gravity. Work is done by the gas on the remaining rocks by raising them against gravity during the expansions, and work is done by the rocks on the gas during the compressions when the remaining rocks are lowered. </p>
<p>This example of the rocks is one of many ways to visualize how to meet the requirement that the cycle be carried out extremely (infinitely) slowly so that the system is always in thermal and mechanical equilibrium with its surroundings. In fact, the usual example is removal of grains of sand, one grain at a time. The Carnot cycle is an idealization to establish the maximum theoretical efficiency of any heat engine operating in a cycle. It is never practically used because it proceeds infinitely slowly. As someone once said, if you put a Carnot engine in your car you would get fantastic fuel economy, but pedestrians would be passing you by!</p>
<blockquote>
<p>Isn't the overall work done still 0?</p>
</blockquote>
<p>Given the explanation of the rocks, and the fact that you said you know how to calculate the work from the diagram, you should see that net work is done in the cycle. The net work done is really the difference between the magnitude of the work done by the gas during the isothermal expansion, minus the magnitude of the work done on the gas during the isothermal compression. The two adiabatic works cancel out.</p>
<blockquote>
<p>Secondly what is the significance of the isothermal adiabatic
expansion / contraction sequence.</p>
</blockquote>
<p>The reversible isothermal expansion is how the system does work on the surroundings by taking in heat isothermally from the surroundings. The isothermal compression allows the system to reject the heat that is always necessary due to the second law. </p>
<p>The reversible adiabatic processes allow the cycle to be completed without needing any heat exchange and change in entropy. The adiabatic expansion brings the system to the lower temperature reservoir <span class="math-container">$T_2$</span> for the isothermal compression to take place. At the end of the isothermal compression the internal energy of the gas is still lower than the starting internal energy. Since all properties, including internal energy, must return to their original values at the completion of the cycle, the adiabatic compression is needed to increase the internal energy back to its original state, without causing any entropy change, since entropy too must return to its original state.</p>
<blockquote>
<p>My previous impression was that any loop in a phase diagram was
reversible if the process was done slowly.</p>
</blockquote>
<p>Yes, provided that in addition there is no friction involved because a process can be carried out slowly and still involve friction.</p>
<blockquote>
<p>Is the efficiency of reversible processes not the same? What is it
about the carnot cycles sequence that gives it the greatest efficiency</p>
</blockquote>
<p>One of the main things that differentiates the Carnot cycle from other reversible cycles is the isothermal expansion and compression. In contrast to any other reversible processes that take in and reject heat, all the heat into the system occurs at the same high temperature, <span class="math-container">$T_1$</span>, and all of the heat out at the same low temperature, <span class="math-container">$T_2$</span>. For other processes the heat exchanges occur at some mean (average) temperature. This maximizes the efficiency of the Carnot cycle by minimizing <span class="math-container">$\frac{T_2}{T_1}$</span> in the equation for efficiency <span class="math-container">$1-\frac{T_2}{T_1}$</span>. This is best visualized by looking at the temperature-entropy diagram.</p>
<p>Hope this helps.</p>
| 285
|
thermodynamics
|
Question about description of Gibbs free energy
|
https://physics.stackexchange.com/questions/511775/question-about-description-of-gibbs-free-energy
|
<p>When introduced to the gibbs free energy, it was derived as follows:</p>
<p>First law: <span class="math-container">$dU=dq+dw$</span></p>
<p>Second law: <span class="math-container">$dS>dq/T$</span> for a spontaneous change.
Note <span class="math-container">$dq$</span> and <span class="math-container">$dw$</span> are inexact differentials.</p>
<p>Subsituting <span class="math-container">$dq=dU-dw$</span>, into the second law gives us:</p>
<p><span class="math-container">$TdS>dU-dw$</span></p>
<p>using <span class="math-container">$dw=-P_{ext}dV$</span></p>
<p><span class="math-container">$Tds>dU+P_{ext}dV$</span></p>
<p>or,</p>
<p><span class="math-container">$dU+P_{ext}dV - TdS<0$</span></p>
<p>Now, keeping pressure and temperature constant, we can say that:
<span class="math-container">$dU+P_{ext}dV - TdS<0$</span><br>
= <span class="math-container">$d(U+P_{ext}V - TS)<0$</span> </p>
<p>= <span class="math-container">$dG<0$</span>, where <span class="math-container">$G$</span> is the gibbs free energy.</p>
<p>Here is my problem.</p>
<p>A few lectures later when we were being introduced to the idea of chemical potential, the gibbs free energy was re written as a function of pressure and temperature in the following way.
<span class="math-container">$dG=Vdp-SdT$</span>, this expression was derived using the result above. My question is that if pressure and temperature were constant in the above expression, isnt <span class="math-container">$dp$</span> and <span class="math-container">$dT$</span> always 0? If so, how is this a valid expression of <span class="math-container">$G$</span>?</p>
|
<p>G is not defined by the equations you wrote. For a pure substance or a mixture of constant chemical composition, it is defined by <span class="math-container">$$G=U+PV-TS$$</span>And this equation applies only to thermodynamic equilibrium states. So, <span class="math-container">$$dG=dU+PdV+VdP-TdS-SdT$$</span>But since <span class="math-container">$dU=TdS-PdV$</span> we are left with <span class="math-container">$$dG=-SdT+VdP$$</span></p>
| 286
|
thermodynamics
|
Can Efficiency of a normal engine in other dimensions be 100%?
|
https://physics.stackexchange.com/questions/513479/can-efficiency-of-a-normal-engine-in-other-dimensions-be-100
|
<p>While I was imagining about carnot engine I got this question whether we can achieve a 100% efficient engine in other dimensions?</p>
|
<p>The efficiency of the most efficient engine or Carnot engine is given by <span class="math-container">$n=1-{Q_2\over Q_1}$</span></p>
<p>For <span class="math-container">$100$</span>% efficiency, <span class="math-container">$Q_2$</span> must be zero which would mean that all heat from the source is converted into heat. This violates the second law of thermodynamics as absolute temperature cannot be negative. This remains unchanged by the number of dimensions in space. Hence the efficiency would be the same. </p>
| 287
|
thermodynamics
|
How to decide whether the process is polytropic, quasistatic, both or neither?
|
https://physics.stackexchange.com/questions/523749/how-to-decide-whether-the-process-is-polytropic-quasistatic-both-or-neither
|
<p>Suppose we have a heat insulated and rigid container with a frictionless piston dividing the container into two parts, the left part being filled with some ideal gas(say mono-atomic), and the right one is vacuum with a spring connected between the right wall and the piston. How do we decide what process the system undergoes now? Will it be a polytropic process? A quastistatic one? Both? Neither? </p>
<p>I think it'll be quasi-static but not polytropic, and as a consequence, the piston's net force will always be zero, but I don't have any explanation for it. How do I explain that to someone? </p>
<p>Also, is it possible to find the heat capacity of this system, considering the heat capacities of the piston, spring etc. negligible for these n moles of ideal gas? How?</p>
|
<p>It won't be polytropic or even quasi-static. The gas will suffer an irreversible expansion, and the final state of the system can be determined.</p>
<p>Assuming that the spring is initially at its unextended length, the force exerted by the spring on the piston during the process will be <span class="math-container">$-k(x-x_0)$</span>, where x is the location relative to the left wall of the cylinder, and <span class="math-container">$x_0$</span> is the its original location. But, if A represents the area of the piston, then <span class="math-container">$-k(x-x_0)=-\frac{k}{A}(V-V_0)$</span> where V is the gas volume at time t, and <span class="math-container">$V_0$</span> is the initial gas volume. So, if we apply Newton's 2nd law of motion to derive a force balance on the piston at any time t, we obtain: <span class="math-container">$$F_g-\frac{k}{A}(V-V_0)=m\frac{d^2x}{dt^2}=\frac{m}{A}\frac{d^2V}{dt^2}$$</span>where <span class="math-container">$F_g$</span> is the force exerted by the gas on the piston at time t and m is the mass of the piston. If we multiply this equation by dV/dt and integrate with respect to time, we can obtain the work done by the gas on the piston (its surroundings) up to time t:
<span class="math-container">$$W_g=\frac{k}{2A}(V-V_0)^2+\frac{m}{2A}\left(\frac{dV}{dt}\right)^2$$</span>In this equation, the first term on the right hand side represents the elastic energy stored in the spring, and the second term represents the kinetic energy of the piston. Even though the piston may oscillate back and forth during this process, eventually, because of viscous damping forces within the gas, the piston will come to rest, and the system will be in its final equilibrium state. At that point, the work done by the gas will be given by:<span class="math-container">$$W_g=\frac{k}{2A}(V_f-V_0)^2$$</span>where <span class="math-container">$V_f$</span> is the final equilibrium gas volume. The final pressure of the gas will then be given by: <span class="math-container">$$P_f=\frac{F_{gf}}{A}=\frac{k}{A^2}(V_f-V_0)\tag{1}$$</span>
Next, applying the first law of thermodynamics to the system, we have <span class="math-container">$$\Delta U=nC_v(T_f-T_0)=-W_{gf}=-\frac{k}{2A}(V_f-V_0)^2$$</span>But, from the ideal gas law, it follows that <span class="math-container">$$nC_v(T_f-T_0)=\frac{C_v}{R}(P_fV_f-P_0V_0)$$</span>Therefore, we have:
<span class="math-container">$$\frac{C_v}{R}(P_fV_f-P_0V_0)=-\frac{k}{2A}(V_f-V_0)^2\tag{2}$$</span>Eqn. 2 can be combined with Eqn. 1 to provide a single quadratic equation for calculating the final volume <span class="math-container">$V_f$</span> (and then the final temperature and final pressure).</p>
| 288
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thermodynamics
|
Heat Exchange Between Two Blocks
|
https://physics.stackexchange.com/questions/530814/heat-exchange-between-two-blocks
|
<p>One question that was posed to me was this: if two otherwise identical blocks are at <span class="math-container">$100$</span> and <span class="math-container">$0$</span> degrees Celsius, and for clarity the hot block is red and the cold block is blue, what is the maximum difference in temperature between the initially cold block and the initially warm block? The catch is you can cut pieces and recombine pieces, but at the end all the blue pieces and red pieces must be back in their original configuration. </p>
<p>Initially, my thought process was that the maximum difference must be <span class="math-container">$0$</span>, since by combining the entirety of both blocks together, you get that each block reaches an equilibrium of <span class="math-container">$50$</span> degrees Celsius, which means the difference is <span class="math-container">$0$</span>.</p>
<p>However, what if you were to combine an infinitesimally small piece of blue with the entirety of the red block, the new temperature would become <span class="math-container">$$T_{n+1}=\lim_{\delta b\to0}(T_n\times\frac{\delta b}{\delta b+r})$$</span></p>
<p>and with <span class="math-container">$N$</span> blue pieces, the final temperature would be
<span class="math-container">$$T_{final} = \lim_{\delta b\to0} T_{initial} \times (\frac{r}{\delta b+r})^N$$</span></p>
<p>If we defined N such that it would be the number of elements of mass <span class="math-container">$\delta b$</span> required to hit the entirety of mass <span class="math-container">$b$</span>, then we could rewrite the limit as
<span class="math-container">$$T_{final} = \lim_{\delta b\to0} T_{initial} \times (\frac{r}{\delta b+r})^{\frac{b}{\delta b}}$$</span></p>
<p>which should become <span class="math-container">$T_{initial}/e$</span>. Why is this the case? Does this violate the second law of thermodynamics? If yes, what did I do wrong?</p>
| 289
|
|
thermodynamics
|
Shape complexity of a substance and its effect on heat retention
|
https://physics.stackexchange.com/questions/533336/shape-complexity-of-a-substance-and-its-effect-on-heat-retention
|
<p>While cooking, I have noticed that a ball of boiled spinach retains its heat for a long time. I understand that spinach is 91% water. It is my hunch that the same volume of water would cool much faster. If 9% of spinach is attributable to other compounds, my first guess might be that water would cool only 9% faster than spinach, but I think spinach would cool at a much slower rate than that. If true, would the complex shape of rolled spinach be the primary factor in its heat retention ability when compared with water? </p>
|
<p>You may need to test your hunch to see if you are right. Water has a very high specific heat, so it is a very good reservoir of heat. A timer and a thermometer might say that it cools off just as slowly.</p>
<p>However, two things might make a "ball of water" cool off more quickly:</p>
<ul>
<li><p>Evaporation. If the whole surface of the ball is exposed to the air (and especially if the water is well above room temperature), then evaporation can cool the surface quickly. The spinach likely inhibits a lot of this by being between the air and most of the water.</p></li>
<li><p>Convection. If the water was just in a pan or a ball, then currents could move warmer interior water to the surface where it could cool. With the spinach in the way, these currents aren't possible. Any hot water in the center has to cool by conduction. This is much slower than if convection were present.</p></li>
</ul>
<p>The shape is probably important, but the complexity of the shape should not be. Just that it inhibits those two processes.</p>
| 290
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thermodynamics
|
Why is $\Delta U = nC_v \Delta T$ true,intuitively, regardless of the path?
|
https://physics.stackexchange.com/questions/550621/why-is-delta-u-nc-v-delta-t-true-intuitively-regardless-of-the-path
|
<p>My question differs from these questions:</p>
<ol>
<li><a href="https://physics.stackexchange.com/questions/336945/when-is-delta-u-nc-v-delta-t-true">When is $\Delta U=nC_V \Delta T$ true?</a></li>
</ol>
<p>Because here he asks to distinguish between <span class="math-container">$C_P$</span> and <span class="math-container">$C_V$</span> and not why it's always <span class="math-container">$C_V$</span></p>
<ol start="2">
<li><a href="https://physics.stackexchange.com/questions/297649/work-done-in-adiabatic-process">Work done in adiabatic process</a></li>
</ol>
<p>Here the answer is "just because it is always the case"</p>
<p>Is there no mathematical proof for this? What is the real physical intuition for this being the case?</p>
|
<p>There are two criteria. First, <span class="math-container">$c_V=nC_V$</span> must be constant; second, we must have that <span class="math-container">$P = f(V) T$</span> for some function <span class="math-container">$f$</span>. Both criteria hold in the specific case of an ideal gas, but neither holds for a general thermodynamical system. I'll give the mathematical explanation first, and then the physical explanation second.</p>
<hr>
<p>Starting from the perspective that <span class="math-container">$U=U(S,V)$</span> and <span class="math-container">$T= \left(\frac{\partial U}{\partial S}\right)_V (S,V)$</span>, note that a small change in <span class="math-container">$S$</span> and <span class="math-container">$V$</span> will cause small changes
<span class="math-container">$$dU=\left(\frac{\partial U}{\partial S}\right)_V dS + \left(\frac{\partial U}{\partial V}\right)_S dV \equiv TdS - PdV$$</span></p>
<p>and
<span class="math-container">$$dT = \left(\frac{\partial T}{\partial S}\right)_V dS + \left(\frac{\partial T}{\partial V}\right)_SdV$$</span></p>
<p>Solving the second equation for <span class="math-container">$dS$</span> and substituting it in the first equation yields</p>
<p><span class="math-container">$$dU = T \frac{1}{\left(\frac{\partial T}{\partial S}\right)_V}dT- \left[T\frac{\left(\frac{\partial T}{\partial V}\right)_S}{\left(\frac{\partial T}{\partial S}\right)_V} + P\right]dV$$</span>
<span class="math-container">$$= c_V dT+ \left[T\frac{\left(\frac{\partial P}{\partial S}\right)_V}{\left(\frac{\partial T}{\partial S}\right)_V} - P\right]dV$$</span></p>
<p>Where we've used that <span class="math-container">$\left(\frac{\partial T}{\partial V}\right)_S = \frac{\partial^2 U}{\partial V\partial S} = -\left(\frac{\partial P}{\partial S}\right)_V$</span>, and that the definition of the specific heat at constant volume is <span class="math-container">$c_V \equiv T \left(\frac{\partial S}{\partial T}\right)_V$</span>. Finally, note that </p>
<p><span class="math-container">$$\frac{\left(\frac{\partial P}{\partial S}\right)_V}{\left(\frac{\partial T}{\partial S}\right)_V} \equiv \left(\frac{\partial P}{\partial T}\right)_V$$</span></p>
<p>so finally
<span class="math-container">$$ dU = c_V dT + \left[T\left(\frac{\partial P}{\partial T}\right)_V - P \right]dV$$</span></p>
<p>Assuming that we are not dealing with variable numbers of particles, what has been written here is completely general, so your question boils down to asking when the second term is zero. The answer is that</p>
<p><span class="math-container">$$\left(\frac{\partial P}{\partial T}\right)_V = \frac{P}{T} \implies P = f(V) T$$</span></p>
<p>for some function <span class="math-container">$V$</span>. If this is the case, then the second term on the right of the preceding equation vanishes, and we have</p>
<p><span class="math-container">$$dU = c_V dT \implies \Delta U = \int c_V dT = \int nC_V dT$$</span>
since <span class="math-container">$C_V$</span> is the specific heat <em>per mole</em>. If <span class="math-container">$C_V$</span> is constant, then this just becomes</p>
<p><span class="math-container">$$\Delta U = nC_V \Delta T$$</span></p>
<hr>
<p>From a physical standpoint, the answer is that the energy of an ideal gas is purely kinetic - the gas particles do not have any long-range interactions with each other at all. As a result, since the temperature can be shown to be a measure of the average kinetic energy of the ideal gas particles, the internal energy of the system is unaffected by changes in volume, as long as the temperature is fixed.</p>
<p>This would not be the case if the particles attracted each other, for example. Putting such a system in a larger box with the same amount of kinetic energy would result in a larger average spacing, and therefore a less negative potential energy (remember that attractive potential energies are negative). Therefore, larger box <span class="math-container">$\implies$</span> more energy, even if the kinetic energy didn't change.</p>
<p>As I showed,
<span class="math-container">$$ dU = c_V dT + \left[T\left(\frac{\partial P}{\partial T}\right)_V - P \right]dV$$</span></p>
<p>The first term on the right describes the change in energy due to change in temperature while holding the volume fixed; the second describes the change in energy due to a change in volume while holding temperature fixed. Because of the lack of interaction between gas particles, the second term goes away, leaving only the first, and so</p>
<p><span class="math-container">$$dU = c_V dT = nC_V dT$$</span></p>
| 291
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thermodynamics
|
Should the thermos flask better be half full or half empty?
|
https://physics.stackexchange.com/questions/551001/should-the-thermos-flask-better-be-half-full-or-half-empty
|
<p>Every evening I am preparing hot water for my two year old son wakes up in the night to get his milk. We use a rather bad isolation can for this. It is a typical metal cylinder shaped can holding half a liter. If I put cooking hot water into it, I know that about 5 hours later it will have room temperature already, but it does the job as my son typically wakes up 2 or three hours after I go to bed, and so he gets his milk temperated.</p>
<p>As I need only about 200ml then to mix up his milk, I was asking myself if it is better to only fill in that amount of hot water or to fill up the whole can.</p>
<p>I guess losing temperature has much to do with the amount of water but also with its surface touching the (colder) room air outside.</p>
<p>With no idea anymore of what my old physics teacher told me twenty five years ago I hope you could share some wisdom for my little story here. Thanks in advance ;)</p>
|
<p>The idealised formula (lumped thermal analysis) for a cooling object, according to Newton's Cooling Law is:</p>
<p><span class="math-container">$$T(t)=T_{\infty}-(T_{\infty}-T_0)\exp\Big(-\frac{t}{\tau}\Big)$$</span></p>
<p>where <span class="math-container">$\tau$</span> is the <em>characteristic time</em>:</p>
<p><span class="math-container">$$\frac{1}{\tau}=\frac{U A}{m c_p}$$</span></p>
<p>with:</p>
<ul>
<li><span class="math-container">$T(t)$</span> is the temperature of the object in time <span class="math-container">$t$</span></li>
<li><span class="math-container">$T_0$</span> is the object's starting temperature and <span class="math-container">$T_{\infty}$</span> the surroundings' temperature</li>
<li><span class="math-container">$A$</span> is the surface area of the object exposed to the surroundings</li>
<li><span class="math-container">$m$</span> is the mass of the object, with specific heat capacity <span class="math-container">$c_p$</span></li>
<li>finally, <span class="math-container">$U$</span> is the <em>overall heat transfer coefficient</em>. <strong>Better quality thermos flasks will generally have lower values of <span class="math-container">$U$</span></strong></li>
</ul>
<p><em>From this we can conclude that for larger <span class="math-container">$m$</span> the rate of cooling will be slower</em>. Note however that greater mass usually also implies greater <span class="math-container">$A$</span>, thereby somewhat offsetting the mass-effect.
<hr>
Following in the footsteps of @probably_someone (comments) I'll explore <span class="math-container">$\frac{A}{m}$</span> for the following shape:</p>
<p><a href="https://i.sstatic.net/2NzEl.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/2NzEl.png" alt="Dewar flask"></a></p>
<p>With <span class="math-container">$m=\rho V$</span> and <span class="math-container">$\rho$</span> (density) a constant we can evaluate <span class="math-container">$\frac{A}{V}$</span> instead:
<span class="math-container">$$V=\frac{\pi D^2}{4}H+\frac12 \frac 43 \pi \Big(\frac{D}{2}\Big)^3=\frac{\pi D^2 H}{4}+\frac{\pi D^3}{12}=\frac{\pi D^2(3H+D)}{12}$$</span></p>
<p><span class="math-container">$$A=\frac12 4\pi \Big(\frac{D}{2}\Big)^2+2 \pi \Big(\frac{D}{2}\Big)H=\frac12 \pi D^2+\pi D H=\frac{\pi D(2H+D)}{2}$$</span></p>
<p>This gives us a result for <span class="math-container">$\frac{A}{V}$</span> of:</p>
<p><span class="math-container">$$\frac{A}{V}=\frac{6(2H+D)}{D(3H+D)}$$</span></p>
<p>Here <span class="math-container">$D$</span> is a constant and only <span class="math-container">$H$</span> increases with <span class="math-container">$m$</span>. Because the coefficient <span class="math-container">$3$</span> in the denominator, instead of <span class="math-container">$2$</span> in the numerator, as <span class="math-container">$H$</span> increases <span class="math-container">$\frac{A}{V}$</span> does indeed decrease which points to lower cooling rate, as expected. </p>
<p><hr>
Another consideration. Assume we half-fill the flask with <em>boiling water</em>. Due to steam formation it is reasonable to expect the entire flask to reach approximately the same temperature. In that case, again as an approximation, the total surface area <span class="math-container">$A$</span> could be used.</p>
| 292
|
thermodynamics
|
Why is heat capacity at const pressure, $C_p$ not a function of volume?
|
https://physics.stackexchange.com/questions/557382/why-is-heat-capacity-at-const-pressure-c-p-not-a-function-of-volume
|
<p>I am reading heat and thermodynamics by Zemansky and while defining heat capacities at constant pressure and volume, it is said that heat capacity at constant pressure is a function of <span class="math-container">$P$</span> and <span class="math-container">$T$</span>. Why not V? and likewise for heat capacity at constant volume is a function of <span class="math-container">$V$</span> and <span class="math-container">$T$</span>, why not <span class="math-container">$P$</span>?</p>
|
<p>The equation of state for a thermodynamical system with a fixed number of particles takes the form
<span class="math-container">$$f(p,V,T)=0$$</span>
for some function <span class="math-container">$f$</span>. For example, in the case of the ideal gas, one has that</p>
<p><span class="math-container">$$f(p,V,T) = pV-nRT = 0$$</span></p>
<p>As a result, <span class="math-container">$p,V,$</span> and <span class="math-container">$T$</span> are not all independent of one another, in the sense that you are not free to specify all three independently.</p>
<hr />
<p>A function like <span class="math-container">$C_p$</span> which is defined for a particular thermodynamic system cannot be considered a function of <span class="math-container">$p,V,$</span> and <span class="math-container">$T$</span> because those three variables cannot be chosen independently. It would be like saying <span class="math-container">$f(x,2x) = 2x^2$</span> is a function of both <span class="math-container">$x$</span> and <span class="math-container">$2x$</span>, which doesn't make sense.</p>
<p>Instead, we work in a framework where we either consider <span class="math-container">$P$</span> to be a function of <span class="math-container">$V$</span> (and other variables) or vice-versa. We can go back and forth between these viewpoints via Legendre transformation. More concretely, when we talk about the internal energy <span class="math-container">$U$</span> or the Helmholtz energy <span class="math-container">$A$</span>, then we are considering <span class="math-container">$p$</span> to be a function of <span class="math-container">$V$</span>. When we talk about the enthalpy <span class="math-container">$H$</span> or the Gibbs energy <span class="math-container">$G$</span>, then we are considering <span class="math-container">$V$</span> to be a function of <span class="math-container">$p$</span>.</p>
<blockquote>
<p>But what about this equation <span class="math-container">$C_p = C_v + \left[\left(\frac{\partial U}{\partial V}\right)_T + P\right] V\beta$</span></p>
</blockquote>
<p>Written out with all of the arguments for the functions, this equation becomes</p>
<p><span class="math-container">$$C_p(P,T) = C_v\big(V(P,T) , T\big) + \left[\left(\frac{\partial U}{\partial V}\right)_T(P,T) + P\right]V(P,T) \beta$$</span></p>
<p>In words, give some <span class="math-container">$P$</span> and <span class="math-container">$T$</span>,</p>
<ol>
<li>First use <span class="math-container">$P,T$</span> to find <span class="math-container">$V(P,T)$</span>. Plug this into the first slot of the function <span class="math-container">$C_v$</span>, and plug <span class="math-container">$T$</span> into the second slot.</li>
<li>Next, calculate <span class="math-container">$\left(\frac{\partial U}{\partial V}\right)_T$</span> and evaluate it at <span class="math-container">$P$</span> and <span class="math-container">$T$</span>. Add <span class="math-container">$P$</span> to the result.</li>
<li>Multiply the output of step 2 by <span class="math-container">$V(P,T)$</span> and <span class="math-container">$\beta$</span></li>
<li>Add the result of step 3 to the result of step 1</li>
</ol>
<p>That is the recipe which defines the function you wrote, and which gives the specific heat at constant pressure as a function of <span class="math-container">$P$</span> and <span class="math-container">$T$</span> only.</p>
| 293
|
thermodynamics
|
Insulating container in real life
|
https://physics.stackexchange.com/questions/559108/insulating-container-in-real-life
|
<p>Is it possible to practically build a perfectly insulating container? If not, what's the best way to build one? By perfectly I mean no heat is trasferred over an infinite amount of time.</p>
|
<p>As Harish Chandra Rajpoot removed his answer instead of rectifying to:</p>
<p>"No, it is not possible to practically build a perfectly insulating container, but even in commercial products we try to get as close as we can get, for example, a "thermos""</p>
<p>The problem is that no material exists that reflects 100% of all electromagnetic radiation, and so, you cannot contain radiation energy dissipation from a closed system within our universe. In other words, every material radiates energy (black body radiation), and it is not possible to contain all this radiation inside one single container. But then we can try our best and use materials that reflect as much as possible back into the thing we want to isolate.</p>
<p>Addressing now the other forms of energy dissipation: convection and conduction. Convection, for liquids and gases, is easy to handle, just build a vacuum around the thing you want to isolate. But then you have conduction. Either you suspend your thing in vacuum, using for example some form of electro-magnetic levitation or you have the container passively supported via physical solid contacts. In the first case, you are wasting energy just to be sure your container stays in no contact with your outer vacuum walls, in the second, contact means there is conduction of heat between your container and the outward vacuum walls, also dissipating heat.</p>
<p>So, in short, we cannot. But the closest we can get, we try, i.e. in refrigerators or thermal-bottles.</p>
| 294
|
thermodynamics
|
How do thermometers actually function?
|
https://physics.stackexchange.com/questions/570445/how-do-thermometers-actually-function
|
<p>I'm trying to figure out what temperature actually is, and there seem to be a lot of answers floating around:</p>
<ul>
<li><a href="https://web.stanford.edu/%7Epeastman/statmech/thermodynamics.html" rel="noreferrer">average kinetic energy per degree of freedom</a></li>
<li><a href="https://physics.stackexchange.com/questions/334004/how-is-temperature-defined-and-measured">a statistical parameter of the Boltzmann distribution</a></li>
<li><a href="https://physics.stackexchange.com/a/567966/165384">partial derivative of energy with respect to entropy</a></li>
<li><a href="https://physics.stackexchange.com/a/567951/165384">a unique parameter that governs the black body radiation an object emits</a></li>
</ul>
<p>and any of these could be fine, except I have no idea how we could possible measure them (except maybe the first, but people generally seem to say that it's actually wrong).</p>
<p>That is, if I take a traditional, thermal-expansion-based thermometer and put it in contact with something, it measures something which corresponds to all of the definitions given above, which is constant at equilibrium, etc. It seems to measure "temperature".</p>
<p>But what is the physical process by which (for example) <span class="math-container">$\frac{\partial E}{\partial S}$</span> governs precisely how much the mercury will rise in a thermometer? Why do we believe that "temperature" (the thing a thermometer measures) corresponds so precisely to "temperature" (any of the four definitions given above)?</p>
<p>(and for that matter, if we take <span class="math-container">$\frac{\partial E}{\partial S}$</span> as a definition of temperature, how do we define <span class="math-container">$S$</span>?)</p>
|
<p>Temperature is a measure based on the most likely value of kinetic energy possessed by the individual particles in a very large ensemble of those particles. The classic "ensemble" in this context is the so-called <em>ideal gas</em>, but the general principle holds for nonideal gases and solids and liquids, with the appropriate corrections which reflect the actual <em>equations of state</em> for those substances.</p>
<p>The kinetic energy of those particles is related to their speed as they vibrate. More vigorous vibrations means more speed, which we measure as an increase in temperature.</p>
<p>That speed increase means the particles exchange greater forces during their collisions, which tends to make them push away harder on one another as the temperature goes up.</p>
<p>This in turn causes the ensemble to expand upon heating.</p>
<p>The liquid mercury in the thermometer is there because its degree of thermal expansion is conveniently large and predictable over a significant temperature range, which makes it a good temperature indicator.</p>
| 295
|
thermodynamics
|
Gas free expansion
|
https://physics.stackexchange.com/questions/593663/gas-free-expansion
|
<blockquote>
<p>It is found that all known
gases cool slightly on undergoing a free expansion. This is consistent with the
kinetic theory idea that temperature is associated with the kinetic energy of
the molecules.
<em>If the gas expands</em>, <strong>then the intermolecular attraction potential
energy goes up as the molecules get further apart.</strong> - Finn thermal</p>
</blockquote>
<p>See the quotation above. I am not sure how does the statement in italic implies the "fact" in bold, that is, why the expansion of a gas increase the potential energy.</p>
<p>Shouldn't the expansion implies the opposite? That is, the molecules gets further apart, and as all potential decrease with distance, shouldn't the potential goes down?</p>
|
<p>"...as all potential decrease with distance..." This is just wrong – sorry.</p>
<p>Imagine that you are pulling further apart the two ends of a rubber band. You have to do work against the tension force and you are thereby transferring energy from your muscles to the rubber band. [The transfer is this way round because the forces you are exerting are in the same direction as the directions in which the forces are moving.]</p>
<p>The situation would be similar if you could pull molecules apart from each other against the attractive Van der Waals (using nano tweezers?) Extra potential energy would be stored in the system of two molecules when pulled apart. In a gas at an ordinary temperature the molecules get further apart in a bigger container because of their random motion – and without the need for tweezers! The extra potential energy comes at the expense of kinetic energy, so in a free expansion the gas cools.</p>
| 296
|
thermodynamics
|
Why is Work and Heat path function? What different paths can they take?
|
https://physics.stackexchange.com/questions/595213/why-is-work-and-heat-path-function-what-different-paths-can-they-take
|
<p>I've seen some interesting analogies explaining why exactly Work and Heat are path function. One where the floors of a building describe the gravitational potential energy, while stairs and elevator are the two paths one can take to reach between the two states of energy.</p>
<p>But my question is how would the same analogy apply to thermodynamics? what would be the different paths taken by the system to go from one state to another? What are the "stairs and elevator" here?</p>
<p>the only guess I can give is that the path is dependent on the volume, pressure, temperature and other state functions, am I right?</p>
|
<blockquote>
<p>the only guess I can give is that the path is dependent on the volume,
pressure, temperature and other state functions, am I right?</p>
</blockquote>
<p>Essentially yes, you are right. To illustrate, see the PV diagram below.</p>
<p>Internal energy (<span class="math-container">$U$</span>), pressure (<span class="math-container">$P$</span>), volume (<span class="math-container">$V$</span>) and temperature (<span class="math-container">$T$</span>) are all thermodynamic properties. For each equilibrium state each of these properties has a unique value.</p>
<p>In the diagram, the two points 1 and 2 are initial and final equilibrium states. The internal energy <span class="math-container">$U$</span> is shown for points 1 and 2. The work done in going from state 1 to 2 is <span class="math-container">$\int_1^2 pdV$</span>, or the area under the path going from state 1 to 2. There are an infinite number of paths (processes) that can connect the two states. Two possible paths are shown, 1-1a-2 and 1-1b-2. Clearly the work done for path 1-1a-2 (the area under the path) is greater than the work done for path 1-1b-2. So work is path dependent. Heat is also path dependent.</p>
<p>Work <span class="math-container">$W$</span> and heat <span class="math-container">$Q$</span> are related to internal energy through the first law which is, for a closed system,</p>
<p><span class="math-container">$$\Delta U=Q-W$$</span></p>
<p>Where <span class="math-container">$Q$</span> is positive if heat is added to the system and work is positive if work is done by the system.</p>
<p>Hope this helps.</p>
<p><a href="https://i.sstatic.net/MgQJ5.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MgQJ5.jpg" alt="enter image description here" /></a></p>
| 297
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thermodynamics
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Why is the knowledge of the thermodynamical state alone is by no means sufficient for the determination of the dynamical state?
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https://physics.stackexchange.com/questions/608496/why-is-the-knowledge-of-the-thermodynamical-state-alone-is-by-no-means-sufficien
|
<p>I am reading the book "Thermodynamics" by Enrico Fermi. There is a passage that goes like this:</p>
<blockquote>
<p>It is evident from what we have said that the knowledge of the
thermodynamical state alone is by no means sufficient for the
determination of the dynamical state. Studying the thermodynamical
state of a homogeneous fluid of given volume at a given temperature
(the pressure is then defined by the equation of state), we observe
that there is an infinite number of states of molecular motion that
correspond to it.With increasing time, the system exists successively
in all these dynamical states that correspond to the given
thermodynamical state. From this point of view we may say that a
thermodynamical state is the ensemble of all the dynamical states
through which, as a result of the molecular motion, the system is
rapidly passing.</p>
</blockquote>
<p>Can someone please help me to understand this? In particular, why is the knowledge of the thermodynamical state alone not sufficient for the determination of the dynamical state?</p>
|
<p>The dynamical state of a (classical) system is a prescription of the position and momentum for each of the constituent particles of the system, while the thermodynamical state of a system is a prescription of the relevant thermodynamical variables like pressure, volume, and temperature. Fermi is simply observing that an infinity of different dynamical states would give rise to the same <span class="math-container">$(P,V,T)$</span>.</p>
| 298
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thermodynamics
|
Phase Transition, Stability of thermodynamic Potentials
|
https://physics.stackexchange.com/questions/629169/phase-transition-stability-of-thermodynamic-potentials
|
<p>I was studying the stability of thermodynamic potentials. It is obvious that internal energy is a convex function of <span class="math-container">$(S,V) $</span>, and in case of instable circumenstances, the system adjust in such a way that the global and local convexity law, is well respected. However, It is mentioned that <span class="math-container">$\frac{\partial u} {\partial s}$</span> must be a continuous function, thus giving us a new form for the <span class="math-container">$u(S)$</span> at constnat <span class="math-container">$V$</span> graph. However, my question is: why it has to be that <span class="math-container">$\frac{\partial u} {\partial s}$</span> is a continuous function? As for the first order phase transitions we can well have some discontinuities of the first partial derivatives of the potentials.</p>
| 299
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