1. Chapter 20
Thermal Properties and Processes
Conceptual Problems
3 • Why is it a bad idea to place a tightly sealed glass bottle that is
completely full of water, into your kitchen freezer in order to make ice?
Determine the Concept Water expands greatly as it freezes. If a sealed glass
bottle full of water is placed in a freezer, as the water freezes there will be no
room for the expansion to take place. The bottle will be broken.
9 •• The phase diagram in Figure 20-15 can be interpreted to yield
information on how the boiling and melting points of water change with altitude.
(a) Explain how this information can be obtained. (b) How might this information
affect cooking procedures in the mountains?
Determine the Concept
(a) With increasing altitude, decreases; from curve OC, the temperature of the
liquid-gas interface decreases as the pressure decreases, so the boiling
temperature decreases. Likewise, from curve OB, the melting temperature
increases with increasing altitude.
(b) Boiling at a lower temperature means that the cooking time will have to be
increased.
13 •• Two solid cylinders made of materials A and B have the same lengths;
their diameters are related by dA = 2dB. When the same temperature difference is
maintained between the ends of the cylinders, they conduct heat at the same rate.
Their thermal conductivities are therefore related by which of the following
equations? (a) kA = kB/4, (b) kA = kB/2, (c) kA = kB, (d) kA = 2kB, (e) kA = 4kB
Picture the Problem The rate at which heat is conducted through a cylinder is
given by xTkAdtdQI ΔΔ== // (see Equation 20-7) where A is the cross-
sectional area of the cylinder.
The heat current in cylinder A is the
same as the heat current in cylinder
B:
BA II =
Substituting for the heat currents
yields: L
T
Ak
L
T
Ak
Δ
=
Δ
BBAA ⇒
A
B
BA
A
A
kk =
409

2. Chapter 20410
Because dA = 2dB:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
B
B
BA
4A
A
kk ⇒ B4
1
A kk =
)(a is correct.
Estimation and Approximation
17 •• You are using a cooking pot to boil water for a pasta dish. The recipe
calls for at least 4.0 L of water to be used. You fill the pot with 4.0 L of room
temperature water and note that this amount of water filled the pot to the brim.
Knowing some physics, you are counting on the volume expansion of the steel pot
to keep all of the water in the pot while the water is heated to a boil. Is your
assumption correct? Explain. If your assumption is not correct, how much water
runs over the sides of the pot due to the thermal expansion of the water?
Determine the Concept The volume of water overflowing is the difference
between the change in volume of the water and the change in volume of the pot.
See Table 20-1 for the coefficient of volume expansion of water and the
coefficient of linear expansion of steel.
Express the volume of water that
overflows when the pot and the water
are heated:
( ) TV
TVTV
VVV
Δ
ΔΔ
ΔΔ
0steelOH
0steel0OH
potOHovefrlow
2
2
2
ββ
ββ
−=
−=
−=
Because the coefficient of volume
expansion of steel is three times its
coefficient of linear expansion:
steelsteel 3αβ =
Substituting for andOH2
β steelβ yields: ( ) TVV Δ3 0steelOHoverflow 2
αα −=
Substitute numerical values and evaluate :overflowV
( )( )( )( ) mL56C20C100L4.0K10113K10207.0 1613
overflow =°−°×−×= −−−−
V
Your assumption was not correct and 56 mL of water overflowed.

3. Thermal Properties and Processes 411
19 •• Estimate the thermal conductivity of human skin.
Picture the Problem We can use the thermal current equation for the thermal
conductivity of the skin. If we model a human body as a rectangular
parallelepiped that is 1.5 m high × 7 cm thick × 50 cm wide, then its surface area
is about 1.8 m2
. We’ll also assume that a typical human, while resting, produces
energy at the rate of 120 W, that normal internal and external temperatures are
33°C and 37°C, respectively, and that an average skin thickness is 1.0 mm.
Use the thermal current equation to
express the rate of conduction of
thermal energy:
I = kA
ΔT
Δx
⇒
x
T
A
I
k
Δ
Δ
=
Substitute numerical values and
evaluate k:
( ) Km
mW
17
m100.1
C33C37
m8.1
W120
3
2 ⋅
=
×
°−°
=
−
k
25 ••• You are in charge of transporting a liver from New York, New York to
Los Angeles, California for a transplant surgery. The liver is kept cold in a
Styrofoam ice chest initially filled with 1.0 kg of ice. It is crucial that the liver
temperature is never warmer than 5.0°C. Assuming the trip from the hospital in
New York to the hospital in Los Angeles takes 7.0 h, estimate the R-value the
Styrofoam walls of the ice chest must have.
Picture the Problem The R factor is the thermal resistance per unit area of a slab
of material. We can use the thermal current equation to express the thermal
resistance of the styrofoam in terms of the maximum amount of heat that can
enter the chest in 7.0 h without raising the temperature above 5.0°C. We’ll
assume that the surface area of the ice chest is 1.0 m2
and that the ambient
temperature is 25°C
The R-factor needed for the
styrofoam walls of the ice chest is the
product of their thermal resistance
and area:
RAR =f (1)
Use the thermal current equation to
express R:
tottot
ΔΔ
Δ
ΔΔ
Q
tT
t
Q
T
I
T
R ===
Substitute for R in equation (1) to
obtain:
tot
f
ΔΔ
Q
tTA
R =

4. Chapter 20412
The total heat entering the chest in
7 h is given by:
OHOHicefice
waterice
warm
ice
melttot
22
ΔTcmLm
QQQ
+=
+=
Substitute for Qtot and simplify to
obtain: ( )OHOHfice
f
22
Δ
ΔΔ
TcLm
tTA
R
+
=
Substitute numerical values and evaluate Rf:
( )
( ) ( )
Btu
fthF
8
C5
Kkg
kJ
18.4
kg
kJ
5.333kg1
Btu
J35.1054
h7
C5
F9
C20
m1029.9
ft1
m1.0 222
2
2
f
⋅⋅°
≈
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
°⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⋅
+
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
°
°
×°⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
×
=
−
R
Thermal Expansion
27 •• You need to fit a copper collar tightly around a steel shaft that has a
diameter of 6.0000 cm at 20ºC. The inside diameter of the collar at that
temperature is 5.9800 cm. What temperature must the copper collar have so that it
will just slip on the shaft, assuming the shaft itself remains at 20ºC?
Picture the Problem Because the temperature of the steel shaft does not change,
we need consider just the expansion of the copper collar. We can express the
required temperature in terms of the initial temperature and the change in
temperature that will produce the necessary increase in the diameter D of the
copper collar. This increase in the diameter is related to the diameter at 20°C and
the increase in temperature through the definition of the coefficient of linear
expansion.
Express the temperature to which the
copper collar must be raised in terms
of its initial temperature and the
increase in its temperature:
TTT Δ+= i
Apply the definition of the
coefficient of linear expansion to
express the change in temperature
required for the collar to fit on the
shaft:
D
DD
D
T
αα
Δ
=
⎟
⎠
⎞
⎜
⎝
⎛ Δ
=Δ
Substitute for ΔT to obtain:
D
D
TT
α
Δ
+= i

5. Thermal Properties and Processes 413
Substitute numerical values and
evaluate T: ( )( )
C220
cm5.9800K1017
cm9800.5cm0000.6
C20 16
°=
×
−
+°= −−
T
The van der Waals Equation, Liquid-Vapor Isotherms, and Phase
Diagrams
33 •• Using Figure 20-16, find the following quantities. (a) The temperature
at which water boils on a mountain where the atmospheric pressure is 70.0 kPa,
(b) the temperature at which water boils in a container where the pressure inside
the container is 0.500 atm, and (c) the pressure at which water boils at 115ºC.
Picture the Problem Consulting Figure 20-16, we see that:
(a) At 70.0 kPa, water boils at approximately C90° .
(b) At 0.500 atm (about 51 kPa), water boils at approximately C78° .
(c) The pressure at which water boils at 115°C is approximately kPa127 .
Conduction
35 • A 20-ft × 30-ft slab of insulation has an R factor of 11. At what rate is
heat conducted through the slab if the temperature on one side is a constant 68ºF
and the temperature of the other side is a constant 30ºF?
Picture the Problem We can use its definition to express the thermal current in
the slab in terms of the temperature differential across it and its thermal resistance
and use the definition of the R factor to express I as a function of ΔT, the cross-
sectional area of the slab, and Rf.
Express the thermal current through
the slab in terms of the temperature
difference across it and its thermal
resistance:
R
T
I
Δ
=
Substitute to express R in terms of
the insulation’s R factor: ff / R
TA
AR
T
I
Δ
=
Δ
=

6. Chapter 20414
Substitute numerical values and
evaluate I:
( )( )( )
h
kBtu
2.1
Btu
Ffth
11
F30F68ft30ft20
2
=
°⋅⋅
°−°
=I
Radiation
41 • The universe is filled with radiation that is believed to be remaining
from the Big Bang. If the entire universe is considered to be a blackbody with a
temperature equal to 2.3 K, what is the λmax (the wavelength at which the power
of the radiation is maximum) of this radiation?
Picture the Problem We can use Wein’s law to find the peak wavelength of this
radiation.
Wein’s law relates the maximum
wavelength of the background
radiation to the temperature of the
universe:
T
Kmm2.898
max
⋅
=λ
Substituting for T gives: mm.31
K3.2
Kmm2.898
max =
⋅
=λ
General Problems
49 •• The solar constant is the power received from the Sun per unit area
perpendicular to the Sun’s rays at the mean distance of Earth from the Sun. Its
value at the upper atmosphere of Earth is about 1.37 kW/m2
. Calculate the
effective temperature of the Sun if it radiates like a blackbody. (The radius of the
Sun is 6.96 × 108
m.).
Picture the Problem We can apply the Stefan-Boltzmann law to express the
effective temperature of the Sun in terms of the total power it radiates. We can, in
turn, use the intensity of the Sun’s radiation in the upper atmosphere of Earth to
approximate the total power it radiates.
Apply the Stefan-Boltzmann law to
relate the energy radiated by the Sun
to its temperature:
4
r ATeP σ= ⇒ 4 r
Ae
P
T
σ
=
Express the surface area of the sun: 2
S4 RA π=

7. Thermal Properties and Processes 415
Relate the intensity of the Sun’s
radiation in the upper atmosphere
to the total power radiated by the
sun:
2
4 R
P
I r
π
= ⇒ IRPr
2
4π=
where R is the Earth-Sun distance.
Substitute for Pr and A in the
expression for T and simplify to
obtain:
4
2
S
2
4
2
S
2
4
4
Re
IR
Re
IR
T
σπσ
π
==
Substitute numerical values and evaluate T:
( )
( ) ( )
K5800
m106.96
Km
W
105.671
m
kW
1.35m101.5
4
28
2
8
2
211
=
×⎟
⎠
⎞
⎜
⎝
⎛
⋅
×
⎟
⎠
⎞
⎜
⎝
⎛
×
=
−
T
53 •• On the average, the temperature of Earth’s crust increases
1.0ºC for every increase in depth of 30 m. The average thermal conductivity of
Earth’s crust material is 0.74 J/m⋅s⋅K. What is the heat loss of Earth per second
due to conduction from the core? How does this heat loss compare with the
average power received from the Sun (which is about 1.37 kW/m2
)?
Picture the Problem We can apply the thermal-current equation to calculate
the heat loss of Earth per second due to conduction from its core. We can
also use the thermal-current equation to find the power per unit area radiated
from Earth and compare this quantity to the solar constant.
Express the heat loss of Earth per
unit time as a function of the thermal
conductivity of Earth and its
temperature gradient:
x
T
kA
dt
dQ
I
Δ
Δ
== (1)
or
x
T
kR
dt
dQ
Δ
Δ
= 2
E4π
Substitute numerical values and evaluate dQ/dt:
( ) kW103.1
m30
C1.0
Ksm
J
0.74m1037.64 1026
×=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ °
⎟
⎠
⎞
⎜
⎝
⎛
⋅⋅
×= π
dt
dQ
Rewrite equation (1) to express the
thermal current per unit area: x
T
k
A
I
Δ
Δ
=

8. Chapter 20416
Substitute numerical values and
evaluate I/A:
( )
2
W/m0.0247
m30
C1.0
KsJ/m0.74
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ °
⋅⋅=
A
I
Express the ratio of I/A to the solar
constant:
%002.0
kW/m1.37
W/m0.0247
constantsolar 2
2
<
=
AI
59 ••• A small pond has a layer of ice 1.00 cm thick floating on its surface.
(a) If the air temperature is –10ºC on a day when there is a breeze, find the rate in
centimeters per hour at which ice is added to the bottom of the layer. The density
of ice is 0.917 g/cm3
. (b) How long do you and your friends have to wait for a
20.0-cm layer to be built up so you can play hockey?
Picture the Problem (a) We can differentiate the expression for the heat that
must be removed from water in order to form ice to relate dQ/dt to the rate of ice
build-up dm/dt. We can apply the thermal-current equation to express the rate at
which heat is removed from the water to the temperature gradient and solve this
equation for dm/dt. In Part (b) we can separate the variables in the differential
equation relating dm/dt and ΔT and integrate to find out how long it takes for a
20.0-cm layer of ice to be built up.
(a) Relate the heat that must be
removed from the water to freeze it
to its mass and heat of fusion:
fmLQ = ⇒
dt
dm
L
dt
dQ
f=
Using the definition of density,
relate the mass of the ice added to
the bottom of the layer to its density
and volume:
AxVm ρρ ==
Differentiate with respect to time
to express the rate of build-up of
the ice:
dt
dx
A
dt
dm
ρ=
Substitute for
dt
dm
to obtain:
dt
dx
AL
dt
dQ
ρf=
The thermal-current equation is:
x
T
kA
dt
dQ Δ
=

9. Thermal Properties and Processes 417
Equate these expressions and
solve for
dt
dx
:
x
T
kA
dt
dx
AL
Δ
=ρf ⇒
x
T
L
k
dt
dx Δ
=
ρf
(1)
Substitute numerical values and
evaluate
dt
dx
:
( )( )
( )( )
cm/h0.70
cm/h0.697
h
s3600
s
m
94.1
m0.0100kg/m917
kg
kJ
333.5
C10C0
Km
W
0.592
3
=
=×=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
°−−°⎟
⎠
⎞
⎜
⎝
⎛
⋅
=
μ
dt
dx
(b) Separating the variables in
equation (1) gives:
dt
L
Tk
xdx
ρf
Δ
=
Integrate x from xi to xf and t′ from
0 to t:
'dt
L
Tk
xdx
tx
x
∫∫
Δ
=
0f
f
i
ρ
and
( ) t
L
Tk
xx
f
2
i
2
f2
1
ρ
Δ
=− ⇒
( )
Tk
xxL
t
Δ
−
=
2
2
i
2
ffρ
Substitute numerical values and evaluate t:
( )( )
( ) ( )[ ]
d12
h24
d1
s3600
h1
s1003.1
m0.010m0.200
C10C0
Km
W
0.5922
kg
kJ
333.5
m
kg
917
6
22
3
=×××=
−
°−−°⎟
⎠
⎞
⎜
⎝
⎛
⋅
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=t