q0/random_k2/random_1.md

Thursday April 9

Topics for this Lecture:
•Thermal Physics

• Temperature & Heat
• Temperature measurement
• Thermal expansion

•Thermal Physics

• Phases & Latent Heat
• Calorimetry

• TF = (9/5)(TC) + 32
• “Temperature” is not “Heat”
• Thermal linear expansion:

• ΔL = αL0ΔT       [α] = K-1

• Heat energy flow:
• Q=c*m*ΔT
• Qlost = Qgained

See the course webpage for slides,
video links, and google docs Q&A links
• Assignment 11 due Friday
• Pre-class due 15min before class
• Help Room: via Teams, 6-9pm
Wed/Thurs, also via email on Wed/Thurs
• SI: via Teams
• Office Hours: via email
(meisel@ohio.edu)

• Phase changes:

Q=m*L

*If your circumstances
hinder your completion of
any of this class’s
assignments, please let
me know
*These slides are
recorded and posted in
advance for your
convenience.
Administrative details
may be out of date. Keep
an eye on your email
please.

Thermal Physics

Why can we only see
warm things in infrared
and not cold objects?

Why do bridges have
these sorts of joints?

How much
energy does it
take to bring
water to a boiling
temperature?

Temperature & Heat
• Temperature:

– A measure of the internal kinetic energy of atoms & molecules within an object.
– This is a property of an object.

• Heat:

– A measure of the transfer of energy between two objects.
– This is a property of an interaction, not of a single object.
– Heat is the energy moving from a high temperature object to a low temperature object.

Temperature Scales

• A few common units are used for temperature:

– Fahrenheit, Celsius, & Kelvin
• Fahrenheit is related to Celsius by:
 TFahrenheit = (9/5)TCelsius + 32
 ΔTFahrenheit = (9/5)ΔTCelsius
• Celsius is related to Kelvin by:
 TKelvin = TCelsius + 273.15

20C is about
room
temperature.

Zero C is
freezing,
10C is cold,
20C is nice,
30C is hot.

• Zero Kelvin is “absolute zero” (lowest possible T)
• Zero Celsius is the freezing point for water,
which is 32 Fahrenheit and 273.15 Kelvin.

• 100 Celsius is the boiling point for water,

which is 212 Fahrenheit and 373.14 Kelvin.

Thermal Expansion of Solids: Linear expansion

• Objects expand when their temperature increases.
• The amount of expansion depends on the material
properties and is known as
the expansion coefficient: α
• α depends on the particular material
(and one the phase of matter). SI unit: K-1
• α describes how much an object expands
in a linear dimension for a given temperature
increase

 ΔL = αL0ΔT

• The fractional change in length is:

 ΔL/L = αΔT
 ΔT must be in Celsius or Kelvin

• Typical α values are (at 20 degrees C):

–Aluminum, Copper, Silver: ~2x10-5 K-1
–Concrete, Glass, Steel : ~1x10-5 K-1
–Diamond: ~0.1x10-5 K-1

This is why you need special
joints for, e.g. bridges

An 0.5m-long rod with a thermal expansion coefficient of α ~ 10-5 K-1 is heated
from room temperature (~293K) to the boiling temperature for water (~373K).
How much extra length does the rod have after heating?

2

(A) 4x10-6 m
(B) 4x10-4 m
(C) 4x10-2 m
(D) 4 m

1. ΔL = αL0ΔT
2. ΔL = (10-5 K-1)(0.5m)(373K - 293K)
3. ΔL = (10-5 K-1)(0.5m)(80K)
4. ΔL = 0.0004m = 400μm This is ~4 human hair widths.

A bi-metallic strip is composed of one metal sandwiched together with
another, so one metal is on top and another metal is on the bottom.
If the metal on the top half has a smaller thermal expansion coefficient
than the metal on the bottom half, which picture best represents what
happens after the bi-metallic strip is heated?

3

(A) 1
(B) 2
(C) 3

1. ΔL = αL0ΔT
2. The top strip has a
smaller α and so it
will expand less than the
bottom strip.

3. To achieve the extra
length on the bottom,
without lengthening the
top-half by the same
amount, the strip is
forced to bend upwards.

Leonard G.

This is how some of the
older spring-based
thermostats work.
Temperature changes cause
the spring to bend,
ultimately completing a
circuit to turn on/off the
heat/AC.

Heat

 Sect 14.1

• Heat is thermal energy that flows from a high-

temperature object to a low-temperature object.

• Heat depends on parameters of a process.

These are so-called “state variables”
– Internal energy (U) of the objects,

i.e. the kinetic energy of atoms/molecules inside

– Temperature (T) of the objects

• Your body radiates heat to the surrounding

environment as light.
• We see that light in the
infrared wavelengths.

Which of these processes will require the most thermal energy?

6

(A) Raising 1kg of water by 10K

(B) Raising 2kg of water by 10K

(C) Raising 2kg of water by 20K

(D) All the same

1. More heat will be required to create a larger temperature change.
2. The more material you have to heat-up, the more energy it will take.

Heat: Heat-energy required to change an object’s temperature

• More heat is required to raise the temperature of a larger object.
• More heat is required to induce a larger temperature change.
• The exact amount of heat will depend on properties of the material.

– These are described the the “specific heat capacity”, c
– c is the amount of energy required to raise the temperature of a given

mass of a particular material by a certain temperature change

– The SI unit of c is: J/(kg*K)
– Typical values of c are ~100J/(kg*K)

Aluminum is 10X higher than this
(which is atypically large for a solid).
Meaning it is hard to heat-up.
Though, liquid water is even higher.

• In a formula:

 Q=c*m*ΔT

• If the temperature of an object decreases,
it will release heat into an environment

• This does not describe phase changes (e.g. liquid water to solid ice)

You want to heat a 1kg block of aluminum and 0.5kg of water by 20K.
Which of these processes will require the most thermal energy?

7

(A) the aluminum block

(B) the water

(C) both the same amount of heat

1. Q= mcΔT
2. Qal= malcalΔT = (1kg)(900J/(kgK))(20K) = 18,000J
3. Qwater= mwatercwaterΔT = (0.5kg)(4000J/(kgK))(20K) = 40,000J
4. The water will require more heat.

caluminum ≈ 900 J/(kg*K)
cwater ≈ 4,000 J/(kg*K)

Calorimetry: Measuring heat-energy transfer via temperature changes

• Suppose I want to measure how much heat-energy a chemical reaction

dumps into a fluid. How do I do it?

• By measuring the temperature before and after the reaction, we get ΔT.
• If we measure the mass of our fluid and know the specific heat,

then the heat energy is given by Q= mcΔT

• The energy “lost” by the chemical reaction is gained by the surrounding

environment:

 Qlost = Qgained

• A key detail is that the stuff you care about (the system),
e.g. the chemical reaction stuff + the water in this picture,
must be thermally isolated from the surrounding environment
(here by coffee cups)

 Sect 14.2

Two liters of water at 353K is added to 1.5-liters of water at 293K.
No energy is lost to the surroundings.
What is the final equilibrium temperature of the water?
Note, 1liter of water has a mass of 1kg.

8

(A) 293 K

(B) 323K

(C) 327K

(D) 353 K

1. Qlost = Qgained
2. (mcΔT)2liters = (mcΔT)1.5liters
3. (2kg)(353K - Tfinal) = (1.5kg)(Tfinal - 293K)
4. 706kg*K - (2kg)Tfinal = (1.5kg)Tfinal - 439.5kg*K
5. 1,145.5kg*K = (3.5kg)Tfinal
6. Tfinal = (1,145.5kg*K)(3.5kg) ≈ 327K

cwater ≈ 4,000 J/(kg*K)

You have two beakers of water, each with 1kg of water (c=4186J/(kgK)) at 293K.
You drop one 353K block of 1kg of metal into each beaker, but one block is made
of aluminum (c= 900J/(kgK)) and the other is made of copper (c= 386J/(kgK)).
Once each beaker reaches thermal equilibrium with the block, which beaker of
water has a higher temperature?

9

(A) beaker with aluminum

(B) beaker with copper

(C) both reach same equilibrium temperature

cwater ≈ 4,000 J/(kg*K)

1. Aluminum has a larger specific heat capacity than copper.
2. This means it takes more energy to change the temperature of aluminum.
3. It also means that more energy is deposited by a change in temperature of the

aluminum block, relative to copper.

Math answer:
1. Qlost = Qgained
2. (mcΔT)block = (mcΔT)water
3. mblockcblock(Tblock - Tfinal) = mwatercwater(Tfinal - Twater)
4. mblockcblockTblock - mblockcblockTfinal = mwatercwaterTfinal - mwatercwaterTwater
5. mblockcblockTblock - mwatercwaterTwater = (mwatercwater + mblockcblock)Tfinal
6. Tfinal = (mblockcblockTblock - mwatercwaterTwater)/(mwatercwater + mblockcblock)

Thermal Energy Storage

• Large objects can be used as energy reservoirs

by storing energy as heat.

• Excess energy can be stored as heat to be
released later, or to smooth-out the power
distribution from a different energy source.
• E.g. store extra solar energy from the day or
wind energy from a windy day to use later
during the night.

• The earth itself is a large hot object

(where the heat comes from radioactive decays
inside the earth’s core).
This heat forms the basis of geothermal energy.

A 1kg block of aluminum (c=900J/(kgK)) has an initial temperature of 300K.
The block is placed over an electric heater which provides 150W of power
for 300 seconds.
After this heating episode, what is the temperature of the aluminum block?

10

(A) 50K

(B) 300K

(C) 350K

(D) 450K

1. Q = mcΔT
2. ΔT = Q/mc
3. Q = P*t = 150W*300s = 45,000J
4. ΔT = (45,000J)/(1kg*900J/(kgK)) = 50K
5. Tfinal = Tinitial + ΔT = 300K + 50K = 350K.

1W = 1J/s

Phases: Definitions

• Solid: Stays the same shape
• Liquid: Takes the shape of the container*, but preserves volume
• Gas: Fills the volume of the container

*To be clear,
cats are not liquids

Phase Changes

• Latent heat describes the amount of energy that needs to be absorbed to change phases

or the amount of energy that is released upon a phase transition
due to breaking/forming molecular bonds

• Energy must be absorbed (“endothermic”) for solid->liquid or liquid->gas
• Energy is released (“exothermic”) for gas->liquid or liquid->solid
• Latent Heat of Fusion, LFusion: solid ↔ liquid
• Latent heat of Vaporization, LVaporization: liquid ↔ gas

Latent Heat

• Latent heat adds/removes heat during a phase transition:
• Following the heat added to bring water from ice to water-vapor:
𝑄𝑄 = 𝑚𝑚𝑚𝑚
Water:
Lfusion = 33.5x104 J/kg
Lvapor = 22.6x105 J/kg
cice = 2090 J/kg ºC
csteam = 2010 J/kg ºC
cwater = 4186 J/kg ºC

QA QB QC

QD

QE

QA=mciceΔT
QB=mLfusion
QC=mcwaterΔT
QD=mLvaporization
QE=mcsteamΔT

How much energy is required to melt 0.5kg of ice?

1

(A) 1045 J

(B) 2090 J

(C) 2093 J

(D) 16.7x104 J

1. This is a phase transition: use Q=mL
Ice to liquid transition: Lfusion = 33.5x104 J/kg
2.
3. Q=(0.5kg)*(33.533.5x104 J/kg) = 16.7x104 J

Calorimetry

• Qlost-from-hot-object  =  Qgained-by-cold-object
• T is positive in this definition; i.e. subtract
the higher temperature from the lower one

• The calorimeter is a classic tool to take

advantage of this

• For the example on the right,

mwatercwaterΔTwater + mcupccupTcup =
mobjectcobjectTobject

Consider 1kg of material A (Blue) and material B (Green).
Which requires the most energy to change from the liquid to the
gas phase?

2

(A) A (B) B (C) They require the same amount

A longer horizontal stretch
corresponds to more added heat
to achieve a phase transition

C

º

p
m
e
T

B

A

gas

liquid

Solid

Heat

A 500g aluminum cylinder at T=20C is heated using steam at T=140C.
What is the minimum mass of the steam required to raise the
temperature of the aluminum cylinder by 40C?

3

(A) 7mg

(B) 0.7kg

(C) 7kg

(D) 70kg

1. Qgained = Qlost
2. Heat Gained by Aluminum = Heat Lost by the Steam (1. Cooling, 2. Condensing, 3. Water Cooling)
3. mAlcAl(ΔT)Al = msteamcsteam(ΔT)steam + msteamLvaporization + mwatercwater(ΔT)water
4. msteam = mwater (note csteam ≠ cwater)
5. mAlcAl(ΔT)Al  = msteam[csteam(ΔT)steam + Lvaporization + cwater(ΔT)water]
6. msteam = mAlcAl(ΔT)Al /[csteam(ΔT)steam + Lvaporization + cwater(ΔT)water]
7. msteam = [(0.5kg)(900J/kgC)(60C-20C)]/[2010J/kgC(140C-100C) +

2.2x106J/kg+ 4186J/kgC(100C-60C)]

8. msteam = 7.18x10-3kg ≈ 7mg

Aluminum:
cAl = 900 J/kg*C

Water:
Lfusion = 33.5x104 J/kg
Lvapor = 22.6x105 J/kg
cice = 2090 J/kg ºC
csteam = 2010 J/kg ºC
cwater = 4186 J/kg ºC

A slug of copper (0.2kg) at 800C is immersed in 0.4kg of water at 80C.
The final temp is 100C. Part of the water boiled off as steam.
What was the mass of the water converted to steam?

4

(A) 7mg

(B) 8mg

(C) 9mg

(D) 10mg

1. Qgained = Qlost
2. Heat lost by copper= Warms-up water + Converts some water to steam
3.mCucCu(ΔT)Cu = mwatercwater(ΔT)water + msteamLvaporization
4.msteam = [(mCucCu(ΔT)Cu) - (mwatercwater(ΔT)water)]/Lvaporization
5.msteam = [(0.2kg)(387J/kgC)(800C-100C) - (0.4kg)(4186J/kgC)(100C-

80C)]/(2.26x106J/kg)

6.msteam = 9.16 x10-3 kg ≈ 9mg  … which is 9mL of liquid water

Copper:
cCu = 387 J/kg*C

Water:
Lfusion = 33.5x104 J/kg
Lvapor = 22.6x105 J/kg
cice = 2090 J/kg ºC
csteam = 2010 J/kg ºC
cwater = 4186 J/kg ºC

It takes 10 minutes for your stove burner to bring 2 quarts of water
(~1.9kg) to boiling (100C) from room temperature (20C).
What is the minimum power output of your stove burner?
(I.e. assume 100% efficiency of heat transfer)

8

(A) 1060 W (B) 600 W (C) 4186 W (D) 640,000 W

1. Power = Energy/Time
2. Energy = Heat = Q = mwatercwaterΔT = (1.9kg)(4186J/kgC)(100C-20C) = 636,272J
3. Time = 10min=600s
4. Power= (636,272J)/(600s) = 1060 J/s = 1060 W

Water:
Lfusion = 33.5x104 J/kg
Lvapor = 22.6x105 J/kg
cice = 2090 J/kgC
csteam = 2010 J/kgC
cwater = 4186 J/kgC
ρwater = 1000kg/m3

You will NOT be asked about the material below on
any homework or exam. Therefore you CAN IGNORE
IT, if you like.

We had to trim the following from this year’s course content due to the pandemic.
However, I think some of you may be curious.

A washer is made of some metal with some thermal coefficient of linear
expansion, α.
If the washer is heated, what happens to R1 and R2?

(A) R1 decreases, R2 increases

(B) R1 increases, R2 increases

(C) R1 increases, R2 decreases

(D) R1 decreases, R2 decreases

1. ΔL = αL0ΔT
2. This means all linear dimensions increase.
3. The radii are linear dimensions, so both will increase.

For instance, consider what would
happen if our washer were made-
up of separate square components.
All sides would lengthen and “R1”
and “R2” would both increase.

Thermal Expansion of Solids: Volume expansion

• Sometimes we want to consider how the volume of an object changes

when we heat it. This is the volume expansion

• Volume expansion: ΔV = βV0ΔT
• The fractional change in volume is: ΔV/V0 = βΔT
• β is the coefficient of volume expansion (SI units: K-1)

– For solids: β≈3α
– For liquids: More complicated, so look-up in table

Most materials expand in volume when
they heat-up. However, water doesn’t.

*Note the change in density is tiny, so we
can usually get away with ignoring it,
until the phase-transition to ice happens.

You fill-up your 12-gallon fuel tank in your sedan up to the brim and
forget to put on your gas cap. Your friend makes the same mistake after
filling-up the 18-gallon fuel tank in her truck to the brim.
It gets really hot outside and the gasoline heats-up.
Who loses the most gas?

5

(A) You do

(B) Your friend does

(C) You both lose the same amount

1. ΔV = βV0ΔT
2. β*ΔT is the same for both of you.
3. But, V0 is greater for the 18-gallon fuel-tank for the truck.
4. So, ΔV will be larger for the truck and it will lose more fuel.