Thermal Physics
Lecture 3
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
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Erwin Sitompul
Thermal Physics 3/1
Chapter 18
First Law of Thermodynamics
Heat Transfer Mechanisms
 Conduction
 Conduction is the transfer of heat from one
substance to another by direct contact.
 Denser substances are better conductors.
The direction of heat is always from the
warmer to the colder substance.
 Consider a slab of face A and thickness L,
whose faces are maintained at temperature
TH and TC by a hot reservoir and a cold
reservoir.
 Let Q be the energy that is transferred as
heat through the slab, from its hot face to its
cold face, in time t.
 Experiment shows that the conduction rate
Pcond (the amount of energy transferred per
unit time) is:
Pcond 
Q
 kA
t
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T H  TC
k
: Thermal conductivity [W/m·K]
L
Erwin Sitompul
Thermal Physics 3/2
Chapter 18
First Law of Thermodynamics
Heat Transfer Mechanisms
 A material that readily transfers energy by
conduction is a good thermal conductor and
has a high value of k.
 For the purpose of insulation, we are more
concerned with poor heat conductors than
with good ones.
 For this reason, the concept of thermal
resistance R is introduced. The R-value of a
slab of thickness L is defined as:
R 
L
k
President University
Erwin Sitompul
Thermal Physics 3/3
Chapter 18
First Law of Thermodynamics
Heat Transfer Mechanisms
 Conduction Through a Composite Slab
 Assume a composite slab, consisting of two
materials having different thickness L1 and
L2 and different thermal conductivities k1
and k2.
 Each face of the slab has area A.
 In the steady state, the conduction rates
through the two materials must be equal.
Let TX be the temperature of the interface
between the two materials, so that:
Pcond  k 2 A
TH  T X
L2
 k1 A
T X  TC
L1
 After a little algebra,
TX 
k1 L 2 TC  k 2 L1T H
k1 L 2  k 2 L1
President University
Erwin Sitompul
Thermal Physics 3/4
Chapter 18
First Law of Thermodynamics
Heat Transfer Mechanisms
 Substituting TX into either equality yields:
Pcond 
A (T H  TC )
L1 k 1  L 2 k 2
 Extending the equation above to apply to any number n of
materials making up a slab:
Pcond 
A ( T H  TC )
n
L
n
kn
i 1
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Erwin Sitompul
Thermal Physics 3/5
Chapter 18
First Law of Thermodynamics
Checkpoint
The figure 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.
ΔTc > ΔTa > ΔTb = ΔTd
 kb and kd ties, then ka, then kc.
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Erwin Sitompul
Thermal Physics 3/6
Chapter 18
First Law of Thermodynamics
Heat Transfer Mechanisms
 Convection
 Convection is the transfer of heat from one substance to another
through the movement of molecules within fluids (liquid or gas).
 As example, convection occurs when water comes in contact with
an object whose temperature is higher than that of the water.
 The temperature of the part of the water that is in contact with the
hot object increases, and (in most cases) water expands and thus
becomes less dense.
 The expanded fluid is now lighter than the surrounding cooler fluid,
and the buoyant forces cause it to rise.
 Some of the surrounding cooler fluid
then flows down to take the place of
the rising warmer fluid, and the
process can
then continue.
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Erwin Sitompul
Thermal Physics 3/7
Chapter 18
First Law of Thermodynamics
Heat Transfer Mechanisms
 Radiation
 Radiation is the transfer of heat through electromagnetic waves or
photons in the form of rays, waves, or particles.
 Radiation does not need a propagating medium and it moves at
the speed of light.
 The heat radiated by the sun can be exchanged between the solar
surface and the Earth’s surface without heating the transitional
space.
 The rate Prad at which an object emits energy via electromagnetic
radiation depends on the object’s surface A and the temperature T
of that area in kelvins and is given by:
Prad   A T
4
σ
ε
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: Stefan-Boltzman constant
= 5.6704×10–8 W/m2·K4
: emissivity of the object’s
surface [0..1]
Erwin Sitompul
Thermal Physics 3/8
Chapter 18
First Law of Thermodynamics
Problem
The cross section of a wall is made of white pine of thickness La and
brick of thickness Ld (=2.0La), sandwiching two layers of unknown
material with identical thicknesses and thermal conductivities.
The thermal conductivity of the pine is ka
and that of the brick is kd (=5.0ka). The
face area A of the wall is unknown.
Thermal conduction through the wall has
reached the steady state; the only known
interface temperatures are T1 = 25°C,
T2 = 20°C, T5 =–10°C. What is the
interface temperature T4?
Pcond,steady state  k a A
ka A
T1  T 2
La
25  20
La
 kd A
Ld
 (5 k a ) A
25  20 
5
2
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T 4  T5
T 4  (  10)
2 La
 T4  (  10) 
 T4   8  C
?If L
= Lc but kb = 2kc,
can you find T3?
Erwin Sitompul
b
Thermal Physics 3/9
Chapter 18
First Law of Thermodynamics
Problem
(a) A cube, 10 cm on a side, of rough steel is heated in a furnace to
a temperature of 400.0°C. If its total emissivity is 0.97,
determine the rate at which it radiates energy from each face.
(b) A black body is a body that completely absorbs all the
electromagnetic radiation falling on it. A small hole in the wall of
an object behaves like a black body. At what rate does radiation
escape from a hole 10 cm2 in area in the wall of a furnace whose
interior is at temperature of 700oC?
(a)
Prad   A T
4
8
2
8
3
 (5.6704  10 )(0.97)(10 )(673.15)
4
 112.94 W
(b)
Prad   A T
4
 (5.6704  10 )(1.00)(10 )(973.15)
 50.85 W
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4
● A black body absorbs
all the radiation
falling on it
Erwin Sitompul
Thermal Physics 3/10
Thermal Physics
Chapter 19
Kinetic Theory
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Erwin Sitompul
Thermal Physics 3/11
Chapter 19
Kinetic Theory
Physics of Gases
 As one of the main subjects in thermodynamics, a gas consists of
atoms or molecules that fill their container’s volume and exert
pressure on the container’s wall.
 There are three variables associated with a gas: volume, pressure,
and temperature. All these can be related with the motion of the
atoms.
 The kinetic theory of gases relates the motion of the atoms to
the volume, pressure, and temperature of the gas.
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Erwin Sitompul
Thermal Physics 3/12
Chapter 19
Kinetic Theory
Avogadro’s Number
 In working with gas consisted of atoms and molecules, it makes
sense to measure the size of the samples in mole.
 By doing so, we can be certain that we are comparing samples
that contain the same number of atoms or molecules.
 One mole is the number of atoms in a 12 g sample of carbon-12.
 The number of atoms or molecules in a mole is determined
experimentally and is given by the Avogadro’s number:
N A  6.02  10 m ol
23
1
 Amedeo Avogadro suggested that all
gases occupying the same volume
under the same conditions of
temperature and pressure contain the
same number of atoms or molecules.
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Erwin Sitompul
Thermal Physics 3/13
Chapter 19
Kinetic Theory
Avogadro’s Number
 The number of moles n contained in a sample of any substance is
equal to the ratio of the number of molecules N in the sample to
the number of molecules NA in 1 mol:
n
N
NA
 We can also find the number of moles n in a sample from the mass
Msam of the sample and either the molar mass M (the mass of 1
mol) or the molecular mass m (the mass of one molecule):
n
M sam

M
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M sam
mN A
Erwin Sitompul
Thermal Physics 3/14
Chapter 19
Kinetic Theory
Ideal Gases
 The objective now is to explain the macroscopic properties of a gas
–pressure, volume, temperature– in terms of the behavior of the
atoms or molecules that make it up.
 Although there are so many different gases, experiments show
that if we confine 1 mol sample of various gases in boxes of
identical volume and hold the gases at the same temperature,
then their measured pressures are nearly the same.
 Further experiments show that, at low enough densities, all real
gases tend to obey the relation:
pV  nR T
p
n
T
R
Ideal Gas Law
: absolute pressure
: number of moles of gas
: temperature (K)
: gas constant
= 8.31 J/mol·K
 An alternative form of the ideal gas law is:
pV  N kT
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N
k
Ideal Gas Law
: number of molecules of gas
: Boltzmann constant
= R/NA
= 1.38×10–23 J/K
Erwin Sitompul
Thermal Physics 3/15
Chapter 19
Kinetic Theory
Ideal Gases
 Work Done by an Ideal Gas at Constant Temperature
(Isothermal Process)
 Suppose we put an ideal gas in a piston-cylinder arrangement.
 Suppose that the gas expands or contracts from an initial volume
Vi to a final volume Vf while we keep the temperature T of the gas
constant.
 Such a process, at constant temperature, is called an isothermal
process.
 On p-V diagram, an isotherm is a curve that connects points that
have the same temperature.
 For n moles of an ideal gas, the equation
of the graph is:
p  nR T
1
V
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 (a constant )
1
V
Erwin Sitompul
Thermal Physics 3/16
Chapter 19
Kinetic Theory
Ideal Gases
 The work done by an ideal gas during an isothermal process is:
Vf
W 

Vf
pdV 
Vi

Vi
nR T
1
V
Vf
dVf  nR T ln V 
Vi
 Completing the calculation:
W  nRT ln
Vf
Vi
Ideal Gas, Isothermal Process
 For an expansion, Vf > Vi
 the natural logarithm of the ratio is positive
 the work W is positive.
 For compression, Vf < Vi
 the natural logarithm of the ratio is negative
 the work W is negative.
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Erwin Sitompul
Thermal Physics 3/17
Chapter 19
Kinetic Theory
Ideal Gases
 Work Done by an Ideal Gas
at Constant Volume (Isochoric Process) and
at Constant Pressure (Isobaric Process)
 To find the work W done by an ideal gas when the temperature
varies, we can go back to the equation:
Vf
W 

pdV
Vi
 For a constant-volume process (isochoric process), then the
equation above yields:
Ideal Gas, Isochoric Process
W 0
 For a constant-pressure process (isobaric process), then the
equation above becomes:
W  p (V f  V i )  p  V
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Ideal Gas, Isobaric Process
Erwin Sitompul
Thermal Physics 3/18
Chapter 19
Kinetic Theory
Checkpoints
An ideal gas has an initial pressure of 3
pressure units and an initial volume of 4
volume units. The table next to the right
gives the final pressure and volume of the
gas (in those same units) in five processes.
Which processes start and end on the same
isotherm?
All but c.
At isothermal process, pV = constant
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Erwin Sitompul
Thermal Physics 3/19
Chapter 19
Kinetic Theory
Problem
A cylinder contains 12 l of oxygen at 20°C and 15 atm. The
temperature is raised to 35°C, and the volume is reduced to 8.5 l.
What is the final pressure of the gas in atmospheres (atm)? Assume
that the gas is ideal.
pV  nR T

pV
 nR  (a constant )
T
p iV i
Ti

p fV f
Tf
p f  pi
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Vi T f
V f Ti
 (15)
(12) (35  273.15)
(8.5) (20  273.15)
Erwin Sitompul
 22.26 atm
Thermal Physics 3/20
Chapter 19
Kinetic Theory
Problem
One mole of oxygen (assume it to be an ideal gas) expands at a
constant temperature T of 310 K from an initial volume Vi of 12 l to a
final volume Vf of 19 l.
How much work is done by the gas during the expansion?
W  nRT ln
Vf
Vi
 (1)(8.31)(310) ln
(19)
(12)
 1183.8 J
?Is there any heat transfer
Q along the expansion?
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Erwin Sitompul
Thermal Physics 3/21
Chapter 19
Kinetic Theory
Class Group Assignments
1. A styrofoam (k = 0.010 W/m2·K) box used to keep drinks cold at a picnic has total
wall area (including the lid) of 0.80 m2 and wall thickness of 2.0 cm. It is filled with
ice and water at 0°C. If the temperature of the outside wall is 30°C, how much ice
melts in one day?
(a) 1.178 kg
(b) 4.210 kg
(c) 2.562 kg
(d) 6.228 kg
(e) 3.114 kg
2. A certain scuba main tank is designed to hold 66 ft3 of air when it is at atmospheric
pressure at 22°C. When the volume of air is compressed to an absolute pressure of
3 000 psi (pound per square inch) and stored in a 10-liter portable tank, the air
becomes so hot that the tank must be allowed to cool before it can be used.
Determine the temperature of the air just after the compression process.
(a) 27°C
(b) 55°C
(c) 40°C
(d) 310°C
(e) 48°C
3. A 2.00-liter container holds half a mole of an ideal gas at a pressure of 12.5 atm. If
R = 8.31 J/mol·K, what is the gas temperature?
(a) 1980 K
(b) 1190 K
(c) 965 K
(d) 609 K
(e) 578 K
4) An air bubble doubles in volume as it rises from the bottom of a lake (ρwater = 1000
kg/m3). Assuming that the process is isothermal, the depth of the lake is:
(a) 21 m
(b) 0.76 m
(c) 4.9 m
(d) 10 m
(e) 0.99
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Erwin Sitompul
Thermal Physics 3/22
Chapter 19
Kinetic Theory
Homework 3
1. (18-57) In part (a) of the next figure, two identical
rectangular rods of metal are welded end to end, with a
temperature of T1 = 0°C on the left side and a temperature
of T2 = 100°C on the right side. In 2.0 min, 10 J is
conducted at a constant rate from the right side to the left
side. How much time would be required to conduct the same
10 J if the rods were welded side to side as in part (b)?
2. (19-13) Air that initially occupies 0.140 m3 at a gauge pressure of 103.0 kPa is
expanded isothermally to an absolute pressure of 101.3 kPa and then cooled at
constant pressure until it reaches its initial volume.
(a) Draw the p-V diagram of the process
(b) Compute the net work done by the air.
(Gauge pressure is the difference between the absolute/actual pressure an
atmospheric pressure.)
 Deadline: Wednesday, 30 April 2014 at 10:30.
 Quiz 1:
Thursday, 1 May 2014 (International Labor Day)
Wednesday, 30 April 2014.
We have make-up class at 10:30-13:00
President University
Erwin Sitompul
Thermal Physics 3/23