Thermal Physics
Eric John M. Jabal
Lesson based on Principles of Physics – Halliday
& Resnick
Thermal Physics
Overview
• Thermometry
• Zeroth Law of
Thermodynamics
• Thermal Expansion
• Calorimetry
• Work and Heat
• First Law of
Thermodynamics
• Heat Transfer
Temperature
Temperature
A scalar quantity measuring hotness
/
Incoldness.
SI units, it is measured on the
Kelvin scale.
Fahrenheit
is based
on
0℉ → freezing point
of brine
90℉ → avg body
temperature
We usually
measure on the
Celsius scale!
0℃ → 𝐟𝐫𝐞𝐞𝐳𝐢𝐧𝐠 𝐩𝐨𝐢𝐧𝐭 of water
100℃ → 𝐛𝐨𝐢𝐥𝐢𝐧𝐠 𝐩𝐨𝐢𝐧𝐭 of water
In Kelvins, 0 K means
absolute zero
temperature!
↓
Lowest possible
temperature
Minimum internal
energy
and300
entropy
While
K means
approximate room
temperature!
Temperature
Temperature Conversions
Celsius
to
Kelvin
𝑇K = 𝑇℃ + 273.15
Celsius
to
Fahrenhe
it
9
𝑇℉ = 𝑇℃ + 32
5
Temperature
Problem
You are a doctor in the US and you
measured the armpit temperature of
a patient to be 100℉.
Is the patient suffering
from a fever, given that
the thermometer readings
for a fever is 37.2℃ or
higher?
Temperature
Problem
Suppose you come across old scientific
notes that describe a temperature scale
called Z on which the boiling point of
water is 65.0°Z and the freezing point is
− 14.0°Z.
To what temperature on the Fahrenheit scale
would a temperature of 𝑇 = −98.0°Z
correspond?
Zeroth Law of
Thermodynamics
“If bodies 𝑨 and 𝑩
are each in thermal
equilibrium with a
third body 𝑻, then 𝑨
and 𝑩 are in thermal
equilibrium with each
other”Pareho init
137.04
K
dalawang
bagay, thermal
equilibrium.
𝐴
𝑇
137.04
K
𝑇
𝐵
Thermal
Expansion
Properties of materials change with
temperature.
Thermal
Linear
Expansion
Expansion
Raising the temperature of a metal rod of
length 𝐿 by ∆𝑇,
Substance
𝜶 (𝟏𝟎−𝟔 /℃)
∆𝐿 ∝ 𝐿∆𝑇
∆𝑳 = 𝜶𝑳∆𝑻
Where 𝛼 is the coefficient of
linear expansion.
𝐿
∆𝑇
∆𝐿
Ice (0℃)
51
Lead
29
Aluminum
23
Brass
19
Copper
17
Concrete
12
Steel
11
Glass (ordinary)
9
Thermal
Linear Expansion
Proble
Expansion
m
An aluminum flagpole is 30 m
high.
By how much does its length
increase as the temperature
increases by 15℃?
Thermal
Volum Expansi
Expansion
e
on
If all dimensions of a solid expand
with temperature,
∆𝑉 ∝ 𝑉∆𝑇
∆𝑽 = 𝜷𝑽∆𝑻
Where 𝛽 is the coefficient of
volume expansion.
This has a relation with 𝛼𝜷 by
= 𝟑𝜶
Unit still in
℃−1 or K −1
Thermal
Are Expansi Proble
Expansion
a
on
m
On the left are four
rectangular metal plates,
with sides 𝐿, 2𝐿, or 3𝐿. They
are all made
of the same
temperature
is to
material, and
their
be increased
by the
same amount.
Rank the plates according to the expected
increase in
(a) their vertical heights and (b) their areas,
Thermal
Volum Expansi Proble
Expansion
e
on
m
On a hot day in Las Vegas, an oil trucker
loaded 37 000 L of diesel fuel. He
encountered cold weather on the way to
Payson, Utah, where the temperature was
23.0 K lower than in Las Vegas, and where
he delivered his entire load. How many
liters did he deliver? The coefficient of
volume expansion for diesel fuel is 9.50 ×
10−4 /℃, and the coefficient of linear
𝑇Room = normal
Calorimetry
Energy transfer
between room and
Coke
This is called
heat Q!
Q is:
𝑇Coke
𝑇Coke==normal
cold
+ if from
environment to
- object
if from
object to
environment
Calorimetry
Heat only exists
Environment
because of
temperature
If 𝑇𝑆 >
𝑄 is
System
differences.
𝑇𝑆
negative
𝑇
𝐸 , 𝑇𝑆 <
If
𝑄 is
𝑄
positive
𝑇
,
If
𝑄 is zero
𝐸 𝑇𝑆 =
𝑇𝐸 ,
Heat is measured in
1
calorie is the amount of heat required to raise the
calories.
temperature of 1 g of water by 1℃
1 cal = 3.968 × 10−3 Btu = 4.1868 J
𝑇𝐸
Calorimetry
Heat Capacity
Amount of heat absorbed/lost per
temperature change.
𝑸
𝑸 = 𝑪∆𝑻
cal/K or
𝑪=
∆𝑻
J/K
Specific Heat
𝑪
Heat capacity per
𝒄=
𝒎
unit mass.
𝑸 = 𝒄𝒎∆𝑻
The specific heat of 𝑐 = 1 cal/g ∙ ℃ = 1 cal/g ∙ K or 4187 J/
water is
kg ∙ K
Calorimetry
Proble
m
What is the (a) heat capacity
and the
(b) heat absorbed by a 70-kg
brass bell exposed to the sun
whose initial temperature is
25℃ and, after some time at
noon, heats up to 60℃? The
specific heat of brass is
given as 380 J/kg ∙ K.
Calorimetry
Problem
An ambient-temperature
stainless steel
teaspoon (25°C) weighing 20 g
is placed in hot coffee (90°C)
also weighing 20 g. If the
specific heat of the teaspoon
is 502.416 J/kg ∙ K and the
specific heat of coffee is the
same as water (4187 J/kg ∙ K),
Calorimetry
Molar Specific
Heat
Heat capacity per
→ amount of substance
23
unit mole 1 mol = 6.02 × 10
where:
elementary units
Avogrado′ s number atoms/molecules
Substanc
e
Specific
Heat
Molar Specific
Heat
cal
g∙K
J
kg ∙ K
J
mol ∙ K
Lead
0.030
5
128
26.5
Tungsten
0.032
1
134
Silver
0.056
4
236
Substanc
e
Specific
Heat
cal
g∙K
J
kg ∙ K
Brass
0.092
380
24.8
Granite
0.19
25.5
Glass
0.20
Substance
Specific
Heat
cal
g∙K
J
kg ∙ K
Mercury
0.03
3
140
790
Ethyl
alcohol
0.58
2430
840
Seawater
0.93
3900
Calorimetry
Proble
m
46.353 moles of an unknown
substance required 1182 J of
heat to raise its temperature
by 1℃. What is its (a) molar
specific heat and its (b)
heat capacity? Round to three
decimal places.
(c) What is the substance
Calorimetry
Heats of
Transformation
Energy required to change the phase of
the material.
𝑸
Per unit mass, the heat of
𝑳=
𝒎
transformation is
The phase change can
either be:
From solid to liquid or → heat of
vice-versa
fusion 𝑳𝑭
From liquid to gas or → heat of
vice-versa
vaporization 𝑳𝑽
Calorimetry
Heats of
Transformation
Melting
Substanc
e
Boiling
Melting
Point (K)
Heat of
Fusion 𝑳𝑭
(kJ/kg)
Boiling
Point (K)
Heat of
Vaporization 𝑳𝑽
(kJ/kg)
Hydrogen
14.0
58.0
20.3
455
Oxygen
54.8
13.9
90.2
213
Mercury
234
11.4
630
296
Water
273
333
373
2256
Lead
601
23.2
2017
858
Silver
1235
105
2323
2336
Copper
1356
207
2868
4730
Calorimetry
Problem
How much heat must be absorbed by ice of
mass m = 720 g at −10℃ to take it to the
liquid state at 15℃?
If we supply the ice
with a total energy
of only 210 kJ (as
heat), what are the
final state and
temperature of the
Work and Heat
Consider a gas in a
cylindrical chamber with a
movable
piston.
If you remove
a few lead
shots from the piston,
allowing the gas to push
with
a
displacement
𝑑
𝑠,
𝑽
𝒇
𝑑𝑊 = 𝐹 ∙ 𝑑 𝑠
𝑾=
𝒑 𝒅𝑽
𝑑𝑊 = (𝑝𝐴)(𝑑𝑠)
𝑑𝑊 = 𝑝(𝐴 𝑑𝑠)
𝑑𝑊 = 𝑝 𝑑𝑉
Insulation
Lead shot
𝐹𝑔
𝑊
Piston
𝐹𝑃
Gas
𝑽𝒊
𝑄
𝒑𝑽 = 𝒏𝑹𝑻
Thermal reservoir
𝑇
Control knob
Work and Heat
Pressure
With a pressure versus
volume graph,
Process
𝑖
𝑊
0
Volume
𝑓
𝑽𝒇
𝑾=
𝒑 𝒅𝑽
𝑽𝒊
Work 𝑊 is path
dependent
When volume
increases, 𝑊 is
positive.
When volume
decreases, 𝑊 is
negative.
First Law of
Thermodynamics
Given an initial and final state, a
system’s work 𝑊 and 𝑄 still depends on
the
process.
However,
it is found experimentally that
𝑸 − 𝑾 is the same for all processes.
∆𝑬𝐢𝐧𝐭 = 𝑸 − 𝑾
𝒅𝑬𝐢𝐧𝐭 = 𝒅𝑸 − 𝒅𝑾
“The internal energy 𝐸𝑖𝑛𝑡 of a system
tends to increase if energy is added as
heat 𝑄 and tends to decrease if energy is
lost as work 𝑊 is done by the system.”
First Law of
Special
Cases
Thermodynamics
Description
Change in
Internal
Energy
Adiabatic
Occurs rapidly so there
is no heat.
∆𝐸int = −𝑊
Constantvolume
Volume is held constant.
∆𝐸int = 𝑄
Cyclical
The system goes back to
initial state.
𝑄=𝑊
Process
First Law of
Problem
Thermodynamics
Let 1.00 kg of liquid water at 100℃ be
converted to steam at 100℃ by boiling at
standard atmospheric pressure.
The volume of water changes from an
initial
valuework
of is done by the system
(a) How much
−3 3
3
1.00
×
10
m
as
a
liquid
to
1.671
m
as
during this process?
(b)
How much energy is transferred as heat
steam.
during the process?
(c) What is the change in the system’s
internal energy?
Heat Transfer
We know that heat is the transfer of energy
between a system and its environment.
But how does that
transfer
There areoccur?
three mechanisms of
heat
transfer.
Conduction,
Conducti
Convecti
Convection,Radiatio
and
Radiation!on
on
n
Heat Transfer
Conduction
Heat makes atoms vibrate. This vibration is
passed along the atoms of the material
through
collisions.
𝑇𝐻
The conduction
rate is
now given
𝑄
𝑇𝐶
𝑃conduction =
𝑡
Given rod with crosssectional area 𝐴 and length
𝑻𝑯 − 𝑻𝑪
𝐿
𝑷𝐜𝐨𝐧𝐝𝐮𝐜𝐭𝐢𝐨𝐧 = 𝜿𝑨
𝑳
𝜅 is the thermal
conductivity
Heat Transfer
𝑳
𝑹=
Conduction
𝜿
We also have the opposite of conductivity:
thermal resistance.
If the conduction
material is made up
of two or more
substances,
𝑨 𝑻𝑯 − 𝑻𝑪
𝑷𝐜𝐨𝐧𝐝𝐮𝐜𝐭𝐢𝐨𝐧 =
𝑳/𝜿
Heat Transfer
Conducti Proble
on
m
Like in the figure below, you hold a 50 cm halfsteel, half-titanium rod to a candle. If the candle
flame has a temperature of 1000℃ while initially
flame
contactof
the rod has a temperature
of is
25℃only
and in
a radius
with the circular face of
2.5 mm, and that the
the rod, (a) what is the
conduction rate of the rod?
The thermal conductivity of
steel is 45 W/m ∙ K while
titanium is 21.9 W/m ∙ K. (b)
Heat Transfer
Conducti Proble
on
m
The figure below shows the face and interface
temperatures of a composite slab consisting of
four materials of identical thicknesses, through
which the heat transfer is steady. Rank the
materials according to their thermal
conductivities, greatest first.
15℃
25℃
10℃
−5.0℃
−10℃
𝑎
𝑏
𝑐
𝑑
Heat Transfer
Convection
Takes advantage of hot fluid rising due to
buoyant force while cold fluid takes its
place.
Heat Transfer
Radiation
Transfer of heat through
electromagnetic
waves.
Called thermal radiation
to differentiate
from
signal
radiation.
The radiation rate of an
object is
𝑷𝐫𝐚𝐝 = 𝝈𝜺𝑨𝑻𝟒
Note
that 𝑇
is only
where 𝜎is the Stefan-Boltzmann
in K
constant equal to 5.6704 ×
−8 W/m2 ∙ K 4
10
and 𝜀 is the
emissivity of the
Heat Transfer
Radiation
Meanwhile, the rate at which an object
absorbs radiation is
𝑷𝐚𝐛𝐬 = 𝝈𝜺𝑨(𝑻𝐞𝐧𝐯 )𝟒
Then, the net rate of
energy exchange due to
thermal radiation is
𝑷𝐧𝐞𝐭 = 𝝈𝜺𝑨 𝑻𝐞𝐧𝐯 𝟒 − 𝑻𝟒
Heat Transfer
Radiati Proble
on
m
A sphere of radius 0.500 m, temperature
27.0℃, and emissivity 0.850 is located in
an environment of temperature 77.0℃.
At what rate does the sphere (a) emit and
(b) absorb thermal radiation?
(c) What is the sphere’s net rate of
energy exchange?
Thermal
Physics
Summary
Thermometry
Temperature
Definition
Measure of
hotness or
coldness
Classifica
Scalar
tion
Temperature Conversions
Celsius
to
Kelvin
𝑇𝐾 = 𝑇𝐶 + 273.15 K
Celsius
to
Fahrenhe
it
9
𝑇𝐹 = 𝑇𝐶 + 32
5
Thermal
Physics
Summary
Law of Thermodynamics
Zeroth Law
Definit Two objects in
thermal
ion
First Law
Definit
ion
Internal energy
increases with
heat but decreases
with work done by
system
equilibrium with
a third object
are also in
thermal
Equatio
∆𝐸int = 𝑄 − 𝑊
equilibrium with
n
First Law Adiabatic Constant
Cyclical
Free
each other
Special
Case
Process
-volume
Process
expansion
s
Equation
∆𝐸int = −𝑊
∆𝐸int = 𝑄
𝑄=𝑊
∆𝐸int = 0
Thermal
Physics
Summary
Thermal Expansion
Type of Thermal
Expansion
Linear Expansion
Volume Expansion
Equation
∆𝐿 = 𝛼𝐿∆𝑇
∆𝑉 = 𝛽𝑉∆𝑇
or
∆𝑉 = 3𝛼𝑉∆𝑇
Thermal
Physics
Summary
Calorimetry
Calorimetry
Parameter
Heat Capacity
Specific Heat
Molar Specific
Heat
𝑄 = 𝑐mol 𝑁∆𝑇
Heat of
Transformation
Fusion 𝑄 = 𝐿𝐹 𝑚
Vaporizat
𝑄 = 𝐿𝑉 𝑚
Equation
𝑄 = 𝐶∆𝑇
𝑄 = 𝑐𝑚∆𝑇
Thermal
Physics
Summary
Heat Transfer
Thermal Mechanism
Conduction
Resistance
Convection
Radiation
Equation
One
material
𝑇𝐻 − 𝑇𝐶
𝑃con = 𝜅𝐴
𝐿
𝐴 𝑇𝐻 − 𝑇𝐶
More
𝑃con =
than one
𝐿/𝜅
𝐿
𝑅=
𝜅
Hot fluid rises, cold fluid falls
Radiation
𝑃rad = 𝜎𝜀𝐴𝑇 4
Absorption
𝑃abs = 𝜎𝜀𝐴(𝑇env )4
Net
𝑃net = 𝜎𝜀𝐴 𝑇env 4 − 𝑇 4
Thermal
Physics
𝑇𝑟
Energy transfer
𝐿
Liquid coating
0℃
Liquid water
0℃
Homewor
k
Liquid water coats an active (growing)
icicle and extends up a short, narrow tube
along the central axis. Because the waterice interface must have a temperature of 0℃,
the water in the tube cannot lose energy
through the sides of the icicle or down
through the tip because there is no
temperature change in those directions. It
can lose energy and freeze only by sending
energy up (through distance 𝐿) to the top of
the icicle, where the temperature 𝑇𝑟 can be
below 0℃. Take 𝐿 = 0.14 m and 𝑇𝑟 = −5℃. Assume
that the central tube and the upward
conduction path both have cross-sectional
area 𝐴. In terms of 𝐴, what rate is (a)
energy conducted upward and (b) mass
Thermal
Physics
Activit
y
Using any AI program / application, generate
one physics problem for each of the following
topics:
•
•
•
•
•
Thermometry
Thermal expansion
Calorimetry
First Law of Thermodynamics
Heat Transfer (Conduction or Radiation)
Then, solve these problems.
Submit your paper before the period ends.