CHAPTER
13 THE TRANSFER OF HEAT
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
(d) The heat conducted during a time t through a bar is given by Q =
( k A ΔT ) t ,
1.
where k is
L
the thermal conductivity, and A and L are the cross-sectional area and length of the bar.
2.
(b) This arrangement conducts more heat for two reasons. First, the temperature difference
ΔT between the ends of each bar is greater in A than in B. Second, the cross-sectional area
available for heat conduction is twice as large in A as in B. A greater cross-sectional area
means more heat is conducted, everything else remaining the same.
3.
4.
(e) This arrangement conducts more heat for two reasons. First, the cross-sectional area A
available for heat flow is twice as large in B than in A. Twice the cross-sectional area means
twice the heat that is conducted. Second, the length of the bars in B is one-half the combined
length in A, which also means that twice the heat is conducted in B as in A. Thus, the heat
conducted in B is 2 × 2 = 4 times greater than that in A.
( k A ΔT ) t . The heat Q,
(e) The heat conducted through a material is given by Q =
L
cross-sectional area A, thickness L, and time t are the same for the three materials. Thus, the
product k ΔT must also be the same for each. Since the temperature difference ΔT across
material 3 is less than that across 2, k3 must be greater than k2. Likewise, since the
temperature difference across material 2 is less than that across 1, k2 must be greater than k1.
5.
k2 = 170 J/ ( s ⋅ m ⋅ C° )
6.
(b) The heat conducted through a material is given by Q =
( k A ΔT ) t
(Equation 13.1). The heat
L
Q, cross-sectional area A, length L, and time t are the same for the two smaller bars. Thus, the
product k ΔT must also be the same for each. Since the thermal conductivity k1 is greater than
k2, the temperature difference ΔT across the left bar is smaller than that across the right bar.
Thus, the temperature where the two bars are joined together (400 °C − ΔT) is greater than
300 °C.
7.
(b) The eagle is being lifted upward by rising warm air. Convection is the method of heat
transfer that utilizes the bulk movement of a fluid, such as air.
Chapter 13 Answers to Focus on Concepts Questions
8.
687
(c) The radiant energy emitted per second is given by Q / t = eσ T 4 A (Equation 13.2). Note
that it depends on the product of T 4 and the surface area A of the cube. The product T 4A is
equal to 1944T04 L20 , 1536T04 L20 , and 864 T04 L20 for B, A, and C, respectively.
9.
Energy emitted per second = 128 J/s
10. (a) The radiant energy emitted per second is given by Q / t = eσ T 4 A (Equation 13.2). The
energy emitted per second depends on the emissivity e of the surface. Since a black surface
has a greater emissivity than a silver surface, the black-painted object emits energy at a
greater rate and, therefore, cools down faster.
11. (d) The radiant energy emitted per second is given by Q / t = eσ T 4 A (Equation 13.2), and it
depends on the product eT 4. Since the energy emitted per second is the same for both
objects, the product eT 4 is the same for both. Since the emissivity of B is 16 times smaller
than the emissivity of A, the temperature of B must be 4 16 = 2 times greater than A.
12. Difference in net powers = 127 W
688 THE TRANSFER OF HEAT
CHAPTER 13 THE TRANSFER OF HEAT
PROBLEMS
______________________________________________________________________________
1.
REASONING Since heat Q is conducted from the blood capillaries to the skin, we can use
( k A ΔT ) t (Equation 13.1) to describe how the conduction process depends
the relation Q =
L
on the various factors. We can determine the temperature difference between the capillaries
and the skin by solving this equation for ΔT and noting that the heat conducted per second
is Q/t.
SOLUTION Solving Equation 13.1 for the temperature difference, and using the fact that
Q/t = 240 J/s, yields
( Q /t) L =
ΔT =
kA
( 240 J/s ) ( 2.0 ×10−3 m )
= 1.5 C°
⎡⎣0.20 J/ ( s ⋅ m ⋅ C° ) ⎤⎦ (1.6 m 2 )
We have taken the thermal conductivity of body fat from Table 13.1.
2.
REASONING AND SOLUTION
a. The heat lost by the oven is
Q=
( kAΔT ) t
=
L
c
hb
gb
0.045 J / (s ⋅ m ⋅ C ° ) 1.6 m 2 160 ° C − 50 ° C 6.0 h
sI
gFGH 3600
J
1h K
0.020 m
= 8.6 × 10 6 J
b. As indicated on the page facing the inside of the front cover, 3.600 × 106 J = 1 kWh, so
that 1 J = 2.78 × 10–7 kWh. Therefore, Q = 2.4 kWh. At $ 0.10 per kWh, the cost is
$ 0.24 .
3.
SSM REASONING AND SOLUTION According to Equation 13.1, the heat per second
lost is
Q k A ΔT [0.040 J/(s ⋅m ⋅ C o )] (1.6 m 2 )(25 C o )
=
=
= 8.0 × 10 2 J/s
–3
t
L
2.0 ×10 m
689
Chapter 13 Problems
where the value for the thermal conductivity k of wool has been taken from Table 13.1.
4.
REASONING The amount of heat Q conducted in a time t is given by
( k AΔT ) t
Q=
(13.1)
L
where k is the thermal conductivity, A is the area, ΔT is the temperature difference, and L is
the thickness. We will apply this relation to each arrangement to obtain the ratio of the heat
conducted when the bars are placed end-to-end to the heat conducted when one bar is placed
on top of the other.
SOLUTION Applying Equation 13.1 for the conduction of heat to both arrangements gives
Qa =
kAa ( ΔT ) t
La
Qb =
and
kAb ( ΔT ) t
Lb
Note that the thermal conductivity k, the temperature difference ΔT, and the time t are the
same in both arrangements. Dividing Qa by Qb gives
kAa ( ΔT ) t
Qa
La
AL
=
= a b
Qb kAb ( ΔT ) t Ab La
Lb
From the text drawing we see that Ab = 2Aa and La = 2Lb. Thus, the ratio is
Qa
Qb
=
Aa Lb
Ab La
=
Aa Lb
( 2 Aa )( 2Lb )
=
1
4
______________________________________________________________________________
5.
SSM REASONING The heat transferred in a time t is given by Equation 13.1,
Q = ( k A ΔT ) t / L . If the same amount of heat per second is conducted through the two
plates, then ( Q / t )al = ( Q / t )st . Using Equation 13.1, this becomes
kal A ΔT
Lal
This expression can be solved for Lst .
=
kst A ΔT
Lst
690 THE TRANSFER OF HEAT
SOLUTION Solving for Lst gives
Lst =
6.
kst
kal
Lal =
14 J/(s ⋅ m ⋅ C°)
(0.035 m) = 2.0 × 10 –3 m
240 J/(s ⋅ m ⋅ C°)
REASONING The heat Q conducted during a time t through a block of length L and
( k A ΔT ) t (Equation 13.1), where k is the thermal
cross-sectional area A is Q =
L
conductivity, and ΔT is the temperature difference.
SOLUTION The cross-sectional area and length of each block are: AA = 2 L20 and
LA = 3L0 , AB = 3L20 and LB = 2 L0 , AC = 6 L20 and LC = L0 . The heat conducted through each
block is:
Case A
QA =
=
2 L2
AA
k ΔT t = 0 k ΔT t =
3L0
LA
2
3
( 23 L0 ) k ΔT t
( 0.30 m ) ⎡⎣250 J/ ( s ⋅ m ⋅ C° )⎤⎦ ( 35 °C − 19 °C )( 5.0 s ) =
4.0 × 103 J
Case B
QB =
=
3L2
AB
k ΔT t = 0 k ΔT t =
LB
2 L0
3
2
( 23 L0 ) k ΔT t
( 0.30 m ) ⎡⎣250 J/ ( s ⋅ m ⋅ C° )⎤⎦ ( 35 °C − 19 °C )( 5.0 s ) =
9.0 × 103 J
Case C
AC
6L20
QC =
k ΔT t =
k ΔT t = ( 6L0 ) k ΔT t
LC
L0
= 6 ( 0.30 m ) ⎡⎣ 250 J/ ( s ⋅ m ⋅ C° )⎤⎦ ( 35 °C − 19 °C )( 5.0 s ) = 3.6 × 104 J
Chapter 13 Problems
7.
691
SSM WWW REASONING AND SOLUTION Values for the thermal conductivities
of Styrofoam and air are given in Table 11.1. The conductance of an 0.080 mm thick sample
of Styrofoam of cross-sectional area A is
ks A
Ls
=
[0.010 J/(s ⋅ m ⋅ C°)] A =
[125 J/(s ⋅ m 2⋅ C°)] A
0.080 × 10−3 m
The conductance of a 3.5 mm thick sample of air of cross-sectional area A is
ka A
La
=
[0.0256 J/(s ⋅ m ⋅ C°)] A =
3.5 × 10
−3
m
[7.3 J/(s ⋅ m 2⋅ C°)] A
Dividing the conductance of Styrofoam by the conductance of air for samples of the same
cross-sectional area A, gives
[125 J/(s ⋅ m 2⋅ C°)] A
= 17
[7.3 J/(s ⋅ m 2⋅ C°)] A
Therefore, the body can adjust the conductance of the tissues beneath the skin by
a factor of 17 .
8.
REASONING The inner radius rin and outer radius rout of the pipe determine the
cross-sectional area A (copper only) of the pipe, which is the difference between the area
Aout of a circle with radius rout and the area Ain of a circle with a radius rin:
2
A = Aout − Ain = π rout
− π rin2
(1)
The heat Q that flows along the pipe in a time t = 15 min is related to the cross-sectional
( kA ΔT ) t (Equation 13.1), where k is the thermal conductivity of copper (see
area A by Q =
L
Table 13.1 in the text), L is the length of the pipe, and ΔT is the temperature difference
between the faucet, where the temperature is 4.0 °C, and the point on the pipe 3.0 m from
the faucet where the temperature is 25 °C: ΔT = 25 D C − 4.0 D C = 21 C D . We will use
Equation (1) and Equation 13.1 to find the inner radius of the pipe.
SOLUTION Solving Equation (1) for the inner radius rin of the pipe yields
2
π rin2 = π rout
−A
or
2
rin2 = rout
−
A
π
or
2
rin = rout
−
A
π
(2)
692 THE TRANSFER OF HEAT
An expression for the cross-sectional area A of the pipe may be obtained by solving
( kA ΔT ) t (Equation 13.1):
Q=
L
QL
(3)
A=
( k ΔT ) t
Substituting Equation (3) into Equation (2) yields
2
−
rin = rout
QL
π ( k ΔT ) t
(4)
Before using Equation (4), we convert the time t from minutes to seconds:
⎛ 60 s
t = 15 min ⎜
⎝ 1 min
(
)
⎞
2
⎟ = 9.0 × 10 s
⎠
The inner radius of the pipe is, therefore,
rin =
( 0.013 m )2 −
(
)⎦ (
)(
π ⎡390 J s ⋅ m ⋅ CD ⎤ 25 DC − 4.0 DC 9.0 × 102 s
⎣
9.
( 270 J )( 3.0 m )
)
= 0.012 m
REASONING The heat Q conducted along the bar is given by the relation Q =
( k A ΔT ) t
L
(Equation 13.1). We can determine the temperature difference between the hot end of the
bar and a point 0.15 m from that end by solving this equation for ΔT and noting that the
heat conducted per second is Q/t and that L = 0.15 m.
SOLUTION Solving Equation 13.1 for the temperature difference, using the fact that
Q/t = 3.6 J/s, and taking the thermal conductivity of brass from Table 13.1, yield
ΔT =
( Q /t)L =
kA
( 3.6 J/s )( 0.15 m )
= 19 C°
⎡⎣110 J/ ( s ⋅ m ⋅ C° ) ⎤⎦ ( 2.6 ×10−4 m 2 )
The temperature at a distance of 0.15 m from the hot end of the bar is
T = 306 °C − 19 C° = 287 °C
Chapter 13 Problems
693
10. REASONING The heat lost by conduction through the wall is Qwall and that lost through
the window is Qwindow. The total heat lost through the wall and window is Qwall + Qwindow.
The percentage of the total heat lost by the window is
⎛
⎞
Qwindow
Percentage = ⎜
⎟ × 100%
⎝ Qwall + Qwindow ⎠
(1)
The amount of heat Q conducted in a time t is given by
Q=
( k AΔT ) t
(13.1)
L
where k is the thermal conductivity, A is the area, ΔT is the temperature difference, and L is
the thickness.
SOLUTION Substituting Equation (13.1) into Equation (1), and letting the symbols “S”
denote the Styrofoam wall and “G” the glass window, we have that
⎛
⎞
Qwindow
Percentage = ⎜
⎟ × 100%
⎝ Qwall + Qwindow ⎠
kG AG ( ΔT ) t
kG AG
⎡
⎤
⎛
⎢
⎥
⎜
LG
LG
⎥ × 100% = ⎜
=⎢
⎢ kS AS ( ΔT ) t kG AG ( ΔT ) t ⎥
⎜ kS AS + kG AG
+
⎢
⎥
⎜ L
LG
LS
LG
⎝ S
⎣
⎦
⎞
⎟
⎟ × 100%
⎟
⎟
⎠
Here we algebraically eliminated the temperature difference ΔT and the time t, since they
are the same in each term. According to Table 13.1 the thermal conductivity of glass is
kG = 0.80 J/ ( s ⋅ m ⋅ C° ) , while the value for Styrofoam is kS = 0.010 J/ ( s ⋅ m ⋅ C° ) . The
percentage of the total heat lost by the window is
694 THE TRANSFER OF HEAT
kG AG
⎛
⎜
LG
Percentage = ⎜
⎜ kS AS + kG AG
⎜ L
LG
⎝ S
⎞
⎟
⎟ × 100%
⎟
⎟
⎠
(
)
⎧
⎡⎣0.80 J/ ( s ⋅ m ⋅ C°) ⎤⎦ 0.16 m 2
⎪
⎪
2.0 × 10−3 m
=⎨
2
⎡⎣0.80 J/ ( s ⋅ m ⋅ C° ) ⎤⎦ 0.16 m2
⎪ ⎡⎣0.010 J/ ( s ⋅ m ⋅ C° ) ⎤⎦ 18 m
+
⎪
0.10 m
2.0 × 10−3 m
⎩
(
)
(
)
⎫
⎪
⎪
⎬ × 100% =
⎪
⎪
⎭
97 %
11. REASONING To find the total heat conducted, we will apply Equation 13.1 to the steel
portion and the iron portion of the rod. In so doing, we use the area of a square for the cross
section of the steel. The area of the iron is the area of the circle minus the area of the square.
The radius of the circle is one half the length of the diagonal of the square.
SOLUTION In preparation for applying Equation 13.1, we need the area of the steel and
the area of the iron. For the steel, the area is simply ASteel = L2, where L is the length of a
side of the square. For the iron, the area is AIron = π R2 – L2. To find the radius R, we use the
Pythagorean theorem, which indicates that the length D of the diagonal is related to the
length of the sides according to D2 = L2 + L2. Therefore, the radius of the circle is
R = D / 2 = 2 L / 2 . For the iron, then, the area is
2
AIron
⎛ 2L ⎞
⎞ 2
2 ⎛π
= π R − L = π ⎜⎜
⎟⎟ − L = ⎜ − 1⎟ L
⎝2 ⎠
⎝ 2 ⎠
2
2
Taking values for the thermal conductivities of steel and iron from Table 13.1 and applying
Equation 13.1, we find
Chapter 13 Problems
695
QTotal = QSteel + QIron
⎡ ( kAΔT ) t ⎤
⎡ ( kAΔT ) t ⎤
⎡
⎛π
⎞ ⎤ ( ΔT ) t
=⎢
+⎢
= ⎢ kSteel L2 + kIron ⎜ − 1⎟ L2 ⎥
⎥
⎥
L
L
⎝2 ⎠ ⎦ L
⎣
⎦Steel ⎣
⎦ Iron ⎣
J
J
⎡⎛
2 ⎛
2⎤
⎞
⎞⎛ π
⎞
= ⎢⎜14
⎟ ( 0.010 m ) + ⎜ 79
⎟ ⎜ − 1⎟ ( 0.010 m ) ⎥
s ⋅ m ⋅ C° ⎠
s ⋅ m ⋅ C° ⎠ ⎝ 2 ⎠
⎝
⎣⎝
⎦
×
( 78 °C − 18 °C )(120 s ) =
0.50 m
85 J
12. REASONING
a. The heat Q conducted through the tile in a time t is given by Q =
( kA ΔT ) t
L
(Equation 13.1), where k is the thermal conductivity of the tile, A is its cross-sectional area,
L is the distance between the outer and inner surfaces, and ΔT is the temperature difference
between the outer and inner surfaces.
b. We will use Q = cm Δ T (Equation 12.4) to find the increase ΔT in the temperature of a
mass m = 2.0 kg of water when an amount of heat Q is transferred to it.
SOLUTION
a. The time t = 5.0 min must be converted to SI units (seconds):
⎛ 60 s
t = 5.0 min ⎜
⎝ 1 min
(
)
⎞
2
⎟ = 3.0 × 10 s
⎠
Because the tile is cubical, its thickness is equal to the length L of one of its sides, and its
cross-sectional area A is the product of two of its side lengths A = (L)(L) = L2. Applying
( kA ΔT ) t (Equation 13.1), we obtain the amount of heat conducted by the tile in five
Q=
L
minutes:
kA ΔT ) t ( kL
(
Q=
=
L
(
2
)
ΔT t
L
)
= ( kL ΔT ) t
(
)(
)
= ⎡ 0.065 J s ⋅ m ⋅ CD ⎤ ( 0.10 m ) 1150 D C − 20.0 D C 3.0 × 10 2 s = 2200 J
⎣
⎦
696 THE TRANSFER OF HEAT
b. Solving Q = cm ΔT (Equation 12.4) for the increase ΔT in temperature, we obtain
ΔT =
Q
cm
(1)
In Equation (1), we will use the value of Q found in part (a), and the specific heat capacity c
of water given in Table 12.2 in the text. With these values, the increase in temperature of
two liters of water is
ΔT =
(
2200 J
)
⎡ 4186 J/ kg ⋅ C ⎤ ( 2.0 kg )
⎣
⎦
D
= 0.26 CD
13. SSM REASONING AND SOLUTION The rate of heat transfer is the same for all three
materials so
Q/t = kpAΔTp/L = kbAΔTb/L = kwAΔTw/L
Let Ti be the inside temperature, T1 be the temperature at the plasterboard-brick interface, T2
be the temperature at the brick-wood interface, and To be the outside temperature. Then
and
kpTi − kpT1 = kbT1 − kbT2
(1)
kbT1 − kbT2 = kwT2 − kwTo
(2)
Solving (1) for T2 gives
T2 = (kp + kb)T1/kb − (kp/kb)Ti
a. Substituting this into (2) and solving for T1 yields
T1 =
( kp /kb ) (1 + kw /kb ) Ti + ( kw /kb ) T0 =
(1 + kw /kb ) (1 + kp /kb ) − 1
21 °C
b. Using this value in (1) yields
T2 = 18 °C
14. REASONING If m kilograms of ice melt in t seconds, then Q = mLf (Equation 12.5) joules
of heat must be delivered to the ice through the copper rod in t seconds, where
Lf = 33.5 ×104 J/kg is the latent heat of fusion of water. The mass of ice per second that
Chapter 13 Problems
melts, then, is given by the ratio
697
m
Q
. The rate
of heat flow through the copper rod is
t
t
Q kA ΔT
=
(Equation 13.1), where k is the thermal conductivity of copper, A
t
L
and L are, respectively, the cross-sectional area and length of the rod, and ΔT = 100.0 C° is
the difference in temperature between the boiling water and the ice-water mixture.
found from
Q
. Dividing this by the
Lf
elapsed time t, we obtain an expression for the mass of ice per second that melts:
SOLUTION Solving Q = mLf (Equation 12.5) for m yields m =
⎛Q⎞
m ⎜⎝ t ⎟⎠
=
t
Lf
Substituting
(1)
Q kA ΔT
=
(Equation 13.1) into Equation (1), we find that
t
L
m kA ΔT ⎡⎣390 J/ ( s ⋅ m ⋅ CD ) ⎤⎦ ( 4.0 ×10−4 m2 )(100.0 CD )
=
=
= 3.1×10−5 kg/s
4
t
Lf L
( 33.5 ×10 J/kg ) (1.5 m )
15. REASONING Heat is delivered from the heating element to the water via conduction. The
amount of heat Q conducted in a time t is given by
Q=
( kcopper AΔT ) t
(13.1)
L
where kcopper is the thermal conductivity of copper, A is the area of the bottom of the pot, ΔT
is the temperature difference, and L is the thickness of the bottom of the pot. Since the water
is boiling under one atmosphere of pressure, the temperature difference is
ΔT = TE – 100.0 °C, where TE is the temperature of the heating element. Substituting this
expression for ΔT into Equation (13.1) and solving for TE, we have
TE = 100.0 °C +
LQ
kCopper At
(1)
When water boils, it changes from the liquid to the vapor phase. The heat required to make
the water change phase is Q = mLv, according to Equation 12.5, where m is the mass and Lv
is the latent heat of vaporization of water.
698 THE TRANSFER OF HEAT
SOLUTION Substituting Q = mLv into Equation (1), and noting that the bottom of the pot
is circular so that its area is A = π R2, we have that
TE = 100.0 °C +
LQ
kCopper At
= 100.0 °C +
L ( mLv )
( )
kCopper π r 2 t
(2)
The thermal conductivity of copper can be found in Table 13.1 ⎡ kcopper = 390 J/ ( s ⋅ m ⋅ C° ) ⎤ ,
⎣
⎦
and the latent heat of vaporization for water can be found in Table 12.3
(Lv = 22.6 × 105 J/kg). The temperature of the heating element is
TE
2.0 × 10−3 m ) ( 0.45 kg ) ( 22.6 × 105 J/kg )
(
= 100.0 °C +
=
⎡⎣390 J/ ( s ⋅ m ⋅ C° ) ⎤⎦ π ( 0.065 m ) (120 s )
2
103.3 °C
16. REASONING
Heat Q flows along the length L of the bar via conduction, so that
( k AΔT ) t , where k is the thermal conductivity of the material
Equation 13.1 applies: Q =
L
from which the bar is made, A is the cross-sectional area of the bar, ΔT is the difference in
temperature between the ends of the bar, and t is the time during which the heat flows. We
will apply this expression twice in determining the length of the bar.
SOLUTION Solving Equation 13.1 for the length L of the bar gives
L=
( k AΔT ) t = k A ( TW − TC ) t
Q
Q
(1)
where TW and TC, respectively are the temperatures at the warmer and cooler ends of the
bar. In this result, we do not know the terms k, A, t, or Q. However, we can evaluate the heat
Q by recognizing that it flows through the entire length of the bar. This means that we can
also apply Equation 13.1 to the 0.13 m of the bar at its cooler end and thereby obtain an
expression for Q:
k A ( T − TC ) t
Q=
D
where the length of the bar through which the heat flows is D = 0.13 m and the temperature
at the 0.13-m point is T = 23 °C, so that ΔT = T − TC . Substituting this result into
Equation (1) and noting that the terms k, A, and t can be eliminated algebraically, we find
Chapter 13 Problems
L=
k A ( TW − TC ) t
Q
=
k A ( TW − TC ) t
k A ( T − TC ) t
=
699
k A ( TW − TC ) t D
k A ( T − TC ) t
D
=
( TW − TC ) D = ( 48 °C − 11 °C )( 0.13 m ) = 0.40 m
23 °C − 11 °C
( T − TC )
17. REASONING The heat Q required to change liquid water at 100.0 °C into steam at
100.0 °C is given by the relation Q = mLv (Equation 12.5), where m is the mass of the water
and Lv is the latent heat of vaporization. The heat required to vaporize the water is
conducted through the bottom of the pot and the stainless steel plate. The amount of heat
( k A ΔT ) t (Equation 13.1), where k is the thermal
conducted in a time t is given by Q =
L
conductivity, A and L are the cross-sectional area and length, and ΔT is the temperature
difference. We will use these two relations to find the temperatures at the aluminum-steel
interface and at the steel surface in contact with the heating element.
SOLUTION
a. Substituting Equation 12.5 into Equation 13.1 and solving for ΔT, we have
ΔT =
QL ( mLv ) L
=
k At
k At
The thermal conductivity kAl of aluminum can be found in Table 13.1, and the latent heat of
vaporization for water can be found in Table 12.3. The temperature difference ΔTAl between
the aluminum surfaces is
ΔTAl
5
−3
mLv ) L ( 0.15 kg ) ( 22.6 ×10 J/kg )( 3.1× 10 m )
(
=
=
= 1.2 C°
kAl At
(
)
⎡⎣ 240 J/ ( s ⋅ m ⋅ C° ) ⎤⎦ 0.015 m 2 ( 240 s )
The temperature at the aluminum-steel interface is TAl-Steel = 100.0 °C + ΔTAl = 101.2 °C .
b. Using the thermal conductivity kss of stainless steel from Table 13.1, we find that the
temperature difference ΔTss between the stainless steel surfaces is
ΔTss
5
−3
mLv ) L ( 0.15 kg ) ( 22.6 ×10 J/kg )(1.4 × 10 m )
(
=
=
= 9.4 C°
kss At
(
)
⎡⎣14 J/ ( s ⋅ m ⋅ C° ) ⎤⎦ 0.015 m 2 ( 240 s )
700 THE TRANSFER OF HEAT
The temperature at the steel-burner interface is T = 101.2 °C + ΔTss = 110.6 °C .
18. REASONING If the cylindrical rod were made of solid copper, the amount of heat it would
conduct in a time t is, according to Equation 13.1, Qcopper = (kcopper A2 ΔT / L)t . Similarly,
the amount of heat conducted by the lead-copper combination is the sum of the heat
conducted through the copper portion of the rod and the heat conducted through the lead
portion:
Qcombination = ⎡ kcopper ( A2 − A1 ) ΔT / L + klead A1ΔT / L ⎤ t .
⎣
⎦
Since the lead-copper combination conducts one-half the amount of heat than does the solid
copper rod, Qcombination = 12 Qcopper , or
kcopper ( A2 − A1 ) ΔT
L
+
klead A1ΔT
L
1 ⎛ kcopper A2 ΔT ⎞
= ⎜
⎟⎟
L
2 ⎜⎝
⎠
This expression can be solved for A1 / A2 , the ratio of the cross-sectional areas. Since the
cross-sectional area of a cylinder is circular, A = π r 2 . Thus, once the ratio of the areas is
known, the ratio of the radii can be determined.
SOLUTION Solving for the ratio of the areas, we have
kcopper
A1
=
A2 2 ( kcopper − klead )
The cross-sectional areas are circular so that A1 / A2 = (π r12 ) /(π r22 ) = (r1 / r2 ) 2 ; therefore,
r1
r2
=
kcopper
2(kcopper − klead )
=
390 J/(s ⋅ m ⋅ C°)
= 0.74
2[390 J/(s ⋅ m ⋅ C°) − 35 J/(s ⋅ m ⋅ C°)]
where we have taken the thermal conductivities of copper and lead from Table 13.1.
Chapter 13 Problems
701
19. SSM WWW REASONING The rate at which heat is conducted along either rod is
given by Equation 13.1, Q / t = ( k A ΔT ) / L . Since both rods conduct the same amount of
heat per second, we have
ks As ΔT
Ls
=
ki Ai ΔT
(1)
Li
Since the same temperature difference is maintained across both rods, we can algebraically
cancel the ΔT terms. Because both rods have the same mass, ms = mi ; in terms of the
densities of silver and iron, the statement about the equality of the masses becomes
ρs ( Ls As ) = ρi ( Li Ai ) , or
As
Ai
=
ρi Li
ρs Ls
(2)
Equations (1) and (2) may be combined to find the ratio of the lengths of the rods. Once the
ratio of the lengths is known, Equation (2) can be used to find the ratio of the cross-sectional
areas of the rods. If we assume that the rods have circular cross sections, then each has an
area of A = π r 2 . Hence, the ratio of the cross-sectional areas can be used to find the ratio of
the radii of the rods.
SOLUTION
a. Solving Equation (1) for the ratio of the lengths and substituting the right hand side of
Equation (2) for the ratio of the areas, we have
Ls
Li
=
ks As
ki Ai
=
ks ( ρi Li )
ki ( ρs Ls )
2
or
⎛ Ls ⎞
ks ρi
⎜⎜ ⎟⎟ =
ki ρs
⎝ Li ⎠
Solving for the ratio of the lengths, we have
Ls
Li
=
ks ρi
ki ρs
=
[420 J/(s ⋅ m ⋅ C°)](7860 kg/m3 )
= 2.0
[79 J/(s ⋅ m ⋅ C°)](10 500 kg/m3 )
b. From Equation (2) we have
π rs2
ρL
= i i
2
ρs Ls
π ri
Solving for the ratio of the radii, we have
2
or
⎛ rs ⎞
ρi Li
⎜⎜ ⎟⎟ =
ρs Ls
⎝ ri ⎠
702 THE TRANSFER OF HEAT
rs
ri
=
ρi ⎛ Li ⎞
7860 kg/m3 ⎛ 1 ⎞
⎜⎜ ⎟⎟ =
⎜
⎟ = 0.61
ρs ⎝ Ls ⎠
10 500 kg/m3 ⎝ 2.0 ⎠
20. REASONING According to Equation 6.10b, power P is the change in energy Q divided by
the time t during which the change occurs, or P = Q/t. The power radiated by a filament is
given by the Stefan-Boltzmann law as
P=
Q
= eσ T 4 A
t
(13.2)
where e is the emissivity, σ is the Stefan-Boltzmann constant, T is the temperature (in
kelvins), and A is the surface area. This expression will be used to find the ratio of the
filament areas of the bulbs.
SOLUTION Solving Equation (13.2) for the area, we have
A=
P
eσ T 4
Taking the ratio of the areas gives
P1
A1
e σT4
= 1 1
P2
A2
e2 σ T24
Setting e2 = e1, and P2 = P1, we have that
P1
4
e1 σ T14
A1
T 4 ( 2100 K )
=
= 24 =
= 0.37
4
A2
P1
T1
2700
K
(
)
4
e1 σ T2
21. SSM WWW REASONING AND SOLUTION Solving the Stefan-Boltzmann law,
Equation 13.2, for the time t, and using the fact that Qblackbody = Qbulb , we have
tblackbody =
Qblackbody
4
σT A
=
Qbulb
4
σT A
=
Pbulb tbulb
σ T4A
Chapter 13 Problems
703
where Pbulb is the power rating of the light bulb. Therefore,
tblackbody =
⎡5.67 ×10
⎣
–8
(100.0 J/s) (3600 s)
J/(s ⋅ m ⋅ K ) ⎤⎦ (303 K) 4 ⎡⎣(6 sides)(0.0100 m) 2 / side ⎤⎦
2
4
⎛ 1 h ⎞⎛ 1 d ⎞
×⎜
⎟⎜
⎟ = 14.5 d
⎝ 3600 s ⎠ ⎝ 24 h ⎠
22. REASONING According to the Stefan-Boltzmann law, the radiant power emitted by the
Q
= e σ T 4 A (Equation 13.2), where Q is the energy radiated in a time t, e is
“radiator” is
t
the emissivity of the surface, σ is the Stefan-Boltzmann constant, T is the temperature in
Kelvins, and A is the area of the surface from which the radiant energy is emitted. We will
apply this law to the “radiator” before and after it is painted. In either case, the same radiant
power is emitted.
SOLUTION Applying the Stefan-Boltzmann law, we obtain the following:
⎛Q⎞
4
= eafter σ Tafter
A
⎜ ⎟
t
⎝ ⎠after
and
⎛Q⎞
4
= ebefore σ Tbefore
A
⎜ ⎟
t
⎝ ⎠before
Since the same radiant power is emitted before and after the “radiator” is painted, we have
⎛Q⎞
⎛Q⎞
=⎜ ⎟
⎜ ⎟
⎝ t ⎠after ⎝ t ⎠before
or
4
4
eafter σ Tafter
A = ebefore σ Tbefore
A
The terms σ and A can be eliminated algebraically, so this result becomes
4
4
eafter σ Tafter
A = ebefore σ Tbefore
A
or
4
4
eafter Tafter
= ebefore Tbefore
Remembering that the temperature in the Stefan-Boltzmann law must be expressed in
Kelvins, so that Tbefore = 62 °C +273 = 335 K (see Section 12.2), we find that
4
Tafter
=
4
ebefore Tbefore
eafter
or
Tafter = 4
ebefore
eafter
0.75
( 335 K ) = 371 K
(Tbefore ) = 4 0.50
On the Celsius scale, this temperature is 371 K − 273 = 98 °C .
704 THE TRANSFER OF HEAT
23. REASONING The radiant energy Q absorbed by the person’s head is given by
Q = e σ T 4 At (Equation 13.2), where e is the emissivity, σ is the Stefan-Boltzmann
constant, T is the Kelvin temperature of the environment surrounding the person
(T = 28 °C + 273 = 301 K), A is the area of the head that is absorbing the energy, and t is the
time. The radiant energy absorbed per second is Q/t = e σ T 4 A.
SOLUTION
a. The radiant energy absorbed per second by the person’s head when it is covered with hair
(e = 0.85) is
(
)
(
)
Q
4
= e σ T 4 A = ( 0.85 ) ⎡5.67 × 10−8 J/ s ⋅ m 2 ⋅ K 4 ⎤ ( 301 K ) 160 × 10−4 m 2 = 6.3 J/s
⎣
⎦
t
b. The radiant energy absorbed per second by a bald person’s head (e = 0.65) is
(
)
(
)
Q
4
= e σ T 4 A = ( 0.65 ) ⎡5.67 × 10−8 J/ s ⋅ m 2 ⋅ K 4 ⎤ ( 301 K ) 160 ×10−4 m 2 = 4.8 J/s
⎣
⎦
t
(
)
24. REASONING The net radiant power of the baking dish is given by Pnet = eσ A T 4 − T04 ,
(Equation 13.3), where e is the emissivity of the dish, σ is the Stefan-Boltzmann constant, A
is the total surface area of the dish, T is the Kelvin temperature of the dish, and T0 is the
temperature of the kitchen. As the dish cools down, both its net radiant power Pnet and
temperature T decrease, but its emissivity e and surface area A remain constant. We will
therefore solve Equation 13.3 for the product of these constant quantities and the StefanBoltzmann constant:
Pnet
eσ A =
(1)
T 4 − T04
(
)
Because the left side of Equation (1) is constant, the ratio on the right hand side cannot
change as the temperature T decreases. We will use this fact to determine the radiant power
Pnet,1 of the baking dish when its temperature is 175 °C.
SOLUTION From Equation (1), we have that
(
Pnet,1
T14 − T04
=
) (
Pnet,2
T24 − T04
)
or
⎡T 4 − T 4 ⎤
Pnet,1 = Pnet,2 ⎢ 14 04 ⎥
⎢⎣ T2 − T0 ⎥⎦
(2)
In Equation (2), Pnet,2 = 12.0 W is the net radiant power when the temperature is 35 °C. We
first convert temperatures to the Kelvin scale by adding 273 to each temperature given in
degrees Celsius (see Equation 12.1). The initial temperature of the baking dish is
Chapter 13 Problems
705
T1 = 175 °C + 273 = 448 K, its final temperature is T2 = 35 °C + 273 = 308 K, and the room
temperature is T0 = 22 °C + 273 = 295 K. Equation (2), then, gives the net radiant power
when the dish is first brought out of the oven:
⎡ ( 448 K )4 − ( 295 K )4 ⎤
⎥ = 275 W
Pnet,1 = (12.0 W ) ⎢
⎢⎣ ( 308 K )4 − ( 295 K )4 ⎥⎦
25. REASONING According to the discussion in Section 13.3, the net power Pnet radiated by
(
)
the person is Pnet = eσ A T 4 − T04 , where e is the emissivity, σ is the Stefan-Boltzmann
constant, A is the surface area, and T and T0 are the temperatures of the person and the
environment, respectively. Since power is the change in energy per unit time (see
Equation 6.10b), the time t required for the person to emit the energy Q contained in the
dessert is t = Q/Pnet.
SOLUTION The time required to emit the energy from the dessert is
t=
Q
Q
=
Pnet eσ A T 4 − T 4
0
(
)
⎛ 4186 J ⎞
The energy is Q = ( 260 Calories ) ⎜
⎟ , and the Kelvin temperatures are
⎝ 1 Calorie ⎠
T = 36 °C + 273 = 309 K and T0 = 21 °C + 273 = 294 K. The time is
t=
( 260 Calories ) ⎛⎜
( 0.75) ⎣⎡5.67 ×10−8
4186 J ⎞
⎟
⎝ 1 Calorie ⎠
= 1.2 ×104 s
4
4
J/ s ⋅ m 2 ⋅ K 4 ⎤ 1.3 m 2 ⎢⎡( 309 K ) − ( 294 K ) ⎥⎤
⎦
⎣
⎦
(
)(
)
26. REASONING The power radiated by an object is given by Q / t = eσ T 4 A (Equation 13.2),
where e is the emissivity of the object, σ is the Stefan-Boltzmann constant, T is the
temperature (in kelvins) of the object, and A is its surface area.
The power that the object absorbs from the room is given by Q/t = eσT04A. Except for the
temperature T0 of the room, this expression has the same form as that for the power radiated
by the object. Note especially that the area A is the surface area of the object, not the room.
Review Example 8 in the text to understand this important point.
706 THE TRANSFER OF HEAT
SOLUTION The object emits three times as much power as it absorbs from the room, so it
follows that (Q/t)emitted = 3(Q/t)absorbed. Using the Stefan-Boltzmann law for each of the
powers, we find
e
σ
T 4
A = 3eσ T04 A
Power emitted
Power absorbed
Solving for the temperature T of the object gives
T = 4 3T0 = 4 3 ( 293 K ) = 386 K
27. SSM REASONING AND SOLUTION The net power generated by the stove is given by
Equation 13.3, Pnet = eσ A ( T 4 − T04 ) . Solving for T gives
F P + T IJ
T =G
H eσ A K
R
= S
T (0.900)[5.67 × 10
1/ 4
net
4
0
7300 W
+ (302 K) 4
–8
2
4
2
J / (s ⋅ m ⋅ K )](2.00 m )
UV
W
1/4
= 532 K
28. REASONING According to the Stefan-Boltzmann law, the power radiated by an object is
Q/t = eσT 4A (Equation 13.2), where e is the emissivity of the object, σ is the
Stefan-Boltzmann constant, T is the temperature (in kelvins) of the object, and A is the
surface area of the object. The surface area of a sphere is A = 4πR2, where R is the radius.
Substituting this expression for A into Equation 13.2, and solving for the radius yields
Q
t
R=
4π e σ T 4
(1)
This expression will be used to find the radius of Sirius B.
SOLUTION Writing Equation (1) for both stars, we have
RSirius
⎛Q⎞
⎜ ⎟
⎝ t ⎠Sirius
=
4
4π eSirius σ TSirius
and
RSun
⎛Q⎞
⎜ ⎟
⎝ t ⎠Sun
=
4
4π eSun σ TSun
Chapter 13 Problems
707
Dividing RSirius by RSun and remembering that (Q/t)Siuris = 0.040(Q/t)Sun and eSirius = eSun,
we obtain
RSirius
=
RSun
⎛Q⎞
⎜ ⎟
⎝ t ⎠Sirius
4
4π eSirius σ TSirius
⎛Q⎞
⎜ ⎟
⎝ t ⎠Sun
4
4π eSun σ TSun
=
⎛Q⎞
0.040 ⎜ ⎟
⎝ t ⎠Sun
4
4π eSun σ TSirius
⎛Q⎞
⎜ ⎟
⎝ t ⎠Sun
4
4π eSun σ TSun
4
0.040TSun
=
4
TSirius
Solving for the radius of Sirius B, and noting that TSirius = 4TSun, gives
2
RSirius
⎛ T
⎞
⎛ T
= 0.040 ⎜ Sun ⎟ RSun = 0.040 ⎜ Sun
⎜T
⎟
⎜ 4T
⎝ Sirius ⎠
⎝ Sun
2
⎞
8
6
⎟⎟ 6.96 × 10 m = 8.7 × 10 m
⎠
(
)
29. SSM REASONING AND SOLUTION The power radiated per square meter by the car
when it has reached a temperature T is given by the Stefan-Boltzmann law, Equation 13.2,
Pradiated / A = eσ T 4 , where Pradiated = Q / t . Solving for T we have
1/ 4
/ A) ⎤
⎡ (P
T = ⎢ radiated
⎥
eσ
⎣
⎦
1/4
⎧
⎫
560 W/m 2
⎪
⎪
=⎨
⎬
–8
2
4
⎪⎩ (1.00) ⎡⎣5.67 ×10 J/(s ⋅ m ⋅ K ) ⎤⎦ ⎪⎭
= 320 K
30. REASONING AND SOLUTION
a. The radiant power lost by the body is
PL = eσ T 4A = (0.80)[5.67 × 10–8 J/(s⋅m2⋅K4)](307 K)4(1.5 m2) = 604 W
The radiant power gained by the body from the room is
Pg = (0.80)[5.67 × 10–8 J/(s⋅m2⋅K4)](298 K)4(1.5 m2) = 537 W
The net loss of radiant power is P = PL − Pg =
b. The net energy lost by the body is
67 W
708 THE TRANSFER OF HEAT
⎛ 1 Calorie ⎞
Q = P t = (67 W)(3600 s) ⎜
⎟ = 58 Calories
⎝ 4186 J ⎠
31. REASONING The liquid helium is at its boiling point, so its temperature does not rise as it
absorbs heat from the radiating shield. Instead, the net heat Q absorbed in a time t converts a
mass m of liquid helium into helium gas, according to Q = mLv (Equation 12.5), where
Lv = 2.1×104 J/kg is the latent heat of vaporization of helium. The ratio of the net heat Q
absorbed by the helium to the elapsed time t is equal to the net power Pnet absorbed by
Q
helium: Pnet = (Equation 6.10b). The net power absorbed by the helium depends upon the
t
temperature T = 77 K maintained by the radiating shield and the temperature T0 = 4.2 K of
(
the boiling helium, as we see from Pnet = eσ A T 4 − T04
)
(Equation 13.3), where
σ = 5.67×10−8 J/(s·m2·K4) is the Stefan-Boltzmann constant, e = 1 is the emissivity of the
container (a perfect blackbody radiator), and A is the surface area of the container. Because
the container is a sphere of radius R, its surface area is given by A = 4πR2. Therefore, the net
power absorbed by the helium can be expressed as
(
)
(
Pnet = eσ A T 4 − T04 = 4π R 2eσ T 4 − T04
)
(1)
SOLUTION Solving Q = mLv (Equation 12.5) for m, we obtain
m=
Solving Pnet =
Q
Lv
(2)
Q
(Equation 6.10b) for Q yields Q = Pnet t . Substituting this into Equation (2),
t
we find that
m=
Q Pnet t
=
Lv
Lv
(3)
Substituting Equation (1) into Equation (3) gives
m=
Pnet t
Lv
=
(
)
4π R 2 eσ T 4 − T04 t
Lv
Since 1 hour is equivalent to 3600 seconds, the mass of helium that boils away in one hour
is
Chapter 13 Problems
709
2
4
4
4π ( 0.30 m ) (1) ⎡⎣5.67 ×10−8 J/ ( s ⋅ m 2 ⋅ K 4 ) ⎤⎦ ⎡⎣( 77 K ) − ( 4.2 K ) ⎤⎦ ( 3600 s )
m=
= 0.39 kg
2.1×104 J/kg
32. REASONING
a. According to Equation 13.2, the radiant power P (or energy per unit time) emitted by an
object is P = Q / t = e σ T 4 A , where e is the emissivity of the object, σ is the
Stefan-Boltzmann constant, T is the temperature (in kelvins) of the object, and A is its
surface area. This expression will allow us to find the ratio of the two absorbed powers. The
reason is that the object and the room have the same constant temperature. Since the
object’s temperature is constant, it must be absorbing the same power that it is emitting.
b. The temperature of the two bars in part (b) of the text drawing can be obtained directly
P
.
from Equation 13.2: T = 4
eσ A
SOLUTION
a. The power absorbed by the two bars in part (b) of the text drawing is given by
Q / t = e σ T 4 A2 (Equation 13.2), where A2 is the total surface area of the two bars that is
exposed to the room: A2 = 28 L20 . The power absorbed by the single bar in part (a) of the
text drawing is Q / t = e σ T 4 A1 , where A1 is the total surface area of the single bar:
A1 = 22 L20 . The ratio of the power P2 absorbed by the two bars to the power P1 absorbed by
the single bar is
(
(
)
)
4
2
P2 eσ T 28 L0
=
= 1.27
P1 eσ T 4 22 L2
0
b. The temperature T2 of the two bars in part (b) of the text drawing and the temperature T1
of the single bar in part (a) are
T2 = 4
P2
e σ A2
and
T1 = 4
Dividing T2 by T1 , and noting that P2 = P1, gives
T2
=
T1
4
4
P2
e σ A2
P1
e σ A1
=4
A1
A2
P1
e σ A1
710 THE TRANSFER OF HEAT
Solving for the temperature of the room and the two bars in part (b) of the text drawing
gives
T2 = T1
4
A1
A2
= ( 450.0 K )
4
22 L20
28 L20
= 424 K
33. SSM REASONING The total radiant power emitted by an object that has a Kelvin
temperature T, surface area A, and emissivity e can be found by rearranging Equation 13.2,
the Stefan-Boltzmann law: Q = eσ T 4 At . The emitted power is P = Q / t = eσ T 4 A .
Therefore, when the original cylinder is cut perpendicular to its axis into N smaller
cylinders, the ratio of the power radiated by the pieces to that radiated by the original
cylinder is
Ppieces eσ T 4 A
2
=
(1)
Poriginal eσ T 4 A1
where A1 is the surface area of the original cylinder, and A2 is the sum of the surface areas
of all N smaller cylinders. The surface area of the original cylinder is the sum of the surface
area of the ends and the surface area of the cylinder body; therefore, if L and r represent the
length and cross-sectional radius of the original cylinder, with L = 10 r ,
A1 = (area of ends) + (area of cylinder body)
= 2(π r 2 ) + (2π r ) L = 2(π r 2 ) + (2π r )(10r ) = 22π r 2
When the original cylinder is cut perpendicular to its axis into N smaller cylinders, the total
surface area A2 is
A2 = N 2(π r 2 ) + (2π r ) L = N 2(π r 2 ) + (2π r )(10r ) = ( 2 N + 20 ) π r 2
Substituting the expressions for A1 and A2 into Equation (1), we obtain the following
expression for the ratio of the power radiated by the N pieces to that radiated by the original
cylinder
Ppieces eσ T 4 A
( 2 N + 20 ) π r 2 = N + 10
2
=
=
11
Poriginal eσ T 4 A1
22π r 2
SOLUTION Since the total radiant power emitted by the N pieces is twice that emitted by
the original cylinder, Ppieces / Poriginal = 2 , we have (N + 10)/11 = 2. Solving this expression
for N gives N = 12 . Therefore, there are 12 smaller cylinders .
Chapter 13 Problems
711
34. REASONING Heat per second is an energy change per unit time, which is power (see
Equation 6.10b). Therefore, the heat per second Qcon/t gained by the sphere due to
conduction is given by Equation 13.1 as
Pcon =
Qcon
t
=
( kArod ΔT )
(13.1)
L
where k is the thermal conductivity of copper (see Table 13.1 in the text), Arod is the
cross-sectional area of the rod, ΔT is the temperature difference between the wall (24 °C)
and the ice (0 °C), and L is the length of the rod outside the sphere. The reason that L in
Equation 13.1 is not equal to the total length of the rod is that the portion of the rod that is
embedded in the ice is at the same temperature as the ice. As there is no temperature
difference across this portion of the rod, there is no heat conduction along it.
The net radiant power gained by the sphere is found from
(
Prad = eσ Asphere T 4 − T04
)
(13.3)
where e is the emissivity of the ice, σ is the Stefan-Boltzmann constant, Asphere is the surface
area of the ice sphere, T is the Kelvin temperature of the room, and T0 is the Kelvin
temperature of the sphere.
SOLUTION One end of the rod is at the center of the ice sphere, so the length L of the rod
outside of the sphere is equal to the total length of the rod minus the sphere’s radius R:
L = 0.25 m − 0.15 m = 0.10 m. With this substitution, Equation 13.1 gives the sphere’s
conductive power gain:
Pcon
D
−4
2
D
D
kArod ΔT ) ⎡⎣390 J ( s ⋅ m ⋅ C ) ⎤⎦ (1.2 × 10 m )( 24 C − 0 C )
(
=
=
= 11 W
0.10 m
L
The surface area of a sphere is given by Asphere = 4πR2, so that the net radiant power gain
given by Equation 13.3 becomes
(
)
(
Prad = eσ Asphere T 4 − T04 = 4π eσ R 2 T 4 − T04
)
(1)
To find the net radiant power gain Prad from Equation (1), we first convert the temperatures
to the Kelvin scale (see Equation 12.1): the temperature of the sphere is
T0 = 0 °C + 273 = 273 K, and the temperature of the room is T = 24 °C + 273 = 297 K.
Substituting the Kelvin temperatures into Equation (1) yields the net radiant power gain of
the sphere:
712 THE TRANSFER OF HEAT
(
Prad = 4π eσ R 2 T 4 − T04
)
(
)
2
4
4
= 4π ( 0.90 ) ⎡5.67 ×10−8 J/ s ⋅ m2 ⋅ K 4 ⎤ ( 0.15 m ) ⎡( 297 K ) − ( 273 K ) ⎤
⎣
⎦
⎣
⎦
= 32 W
Thus, the ratio of the heat gain per second due to conduction to the net heat gain per second
due to radiation is
Pcon
Prad
=
11 W
= 0.34
32 W
35. SSM REASONING The heat conducted through the iron poker is given by Equation
13.1, Q = ( kA ΔT ) t / L . If we assume that the poker has a circular cross-section, then its
cross-sectional area is A = π r 2 . Table 13.1 gives the thermal conductivity of iron as
79 J / (s ⋅ m ⋅ C ° ) .
SOLUTION The amount of heat conducted from one end of the poker to the other in 5.0 s
is, therefore,
b
g c
79 J / s ⋅ m ⋅ C ° π 5.0 × 10 –3 m
( k A ΔT ) t
Q=
=
L
1.2 m
36. REASONING AND SOLUTION
refrigerator walls is
h b502 ° C – 26 ° C gb5.0 sg =
2
12 J
The rate at which energy is gained through the
(
)
2
Q kA ΔT [ 0.030 J/(s ⋅ m ⋅ C°) ] 5.3 m ( 25 °C − 5 °C )
=
=
= 42 J/s
t
L
0.075 m
Therefore, the amount of heat per second that must be removed from the unit to keep it cool
is 42 J / s .
37. REASONING The radiant energy Q radiated by the sun is given by Q = e σ T 4 At
(Equation 13.2), where e is the emissivity, σ is the Stefan-Boltzmann constant, T is its
temperature (in Kelvins), A is the surface area of the sun, and t is the time. The radiant
energy emitted per second is Q/t = e σ T 4 A. Solving this equation for T gives the surface
temperature of the sun.
Chapter 13 Problems
713
SOLUTION The radiant power produced by the sun is Q/t = 3.9 × 1026 W. The surface area
of a sphere of radius r is A = 4πr2. Since the sun is a perfect blackbody, e = 1. Solving
Equation 13.2 for the surface temperature of the sun gives
T=
4
Q/t
3.9 × 1026 W
=
e σ 4π r 2 4 (1) ⎡5.67 × 10−8 J/ s ⋅ m 2 ⋅ K 4 ⎤ 4π 6.96 × 108 m
⎣
⎦
(
)
(
)
2
= 5800 K
38. REASONING The net rate at which energy is being lost via radiation cannot exceed the
production rate of 115 J/s, if the body temperature is to remain constant. The net rate at
which an object at temperature T radiates energy in a room where the temperature is T0 is
given by Equation 13.3 as Pnet = eσA(T4 – T04). Pnet is the net energy per second radiated.
We need only set Pnet equal to 115 J/s and solve for T0. We note that the temperatures in this
equation must be expressed in Kelvins, not degrees Celsius.
SOLUTION According to Equation 13.3, we have
(
Pnet = eσ A T 4 − T04
)
or
T04 = T 4 −
Pnet
eσ A
Using Equation 12.1 to convert from degrees Celsius to Kelvins, we have T = 34 + 273 =
307 K. Using this value, it follows that
T0 = 4 T 4 −
Pnet
eσ A
= 4 ( 307 K ) −
4
115 J/s
(
)(
0.700 ⎡5.67 × 10−8 J/ s ⋅ m 2 ⋅ K 4 ⎤ 1.40 m 2
⎣
⎦
)
= 287 K (14 °C)
39. REASONING AND SOLUTION The heat Q conducted during a time t through a wall of
thickness L and cross sectional area A is given by Equation 13.1:
Q=
kA ΔT t
L
The radiant energy Q, emitted in a time t by a wall that has a Kelvin temperature T, surface
area A, and emissivity e is given by Equation (13.2):
714 THE TRANSFER OF HEAT
Q = eσ T 4 At
If the amount of radiant energy emitted per second per square meter at 0 °C is the same as
the heat lost per second per square meter due to conduction, then
⎛Q⎞
⎛Q⎞
=⎜ ⎟
⎜ ⎟
⎝ t A ⎠conduction ⎝ t A ⎠ radiation
Making use of Equations 13.1 and 13.2, the equation above becomes
k ΔT
= eσ T 4
L
Solving for the emissivity e gives:
e=
[1.1 J/(s ⋅ m ⋅ K)](293.0 K − 273.0 K)
k ΔT
=
= 0.70
4
(0.10 m)[5.67 ×10−8 J/(s ⋅ m 2 ⋅ K 4 )] (273.0 K)4
Lσ T
Remark on units: Notice that the units for the thermal conductivity were expressed as
J/(s.m.K) even though they are given in Table 13.1 as J/(s.m.C°). The two units are
equivalent since the "size" of a Celsius degree is the same as the "size" of a Kelvin; that is,
1 C° = 1 K. Kelvins were used, rather than Celsius degrees, to ensure consistency of units.
However, Kelvins must be used in Equation 13.2 or any equation that is derived from it.
40. REASONING AND SOLUTION According to Equation 13.2, for the sphere we have
Q/t = eσAsTs4, and for the cube Q/t = eσAcTc4. Equating and solving we get
Tc4 = (As/Ac)Ts4
Now
As/Ac = (4π R2)/(6L2)
1/ 3
⎛ 3 ⎞
The volume of the sphere and the cube are the same, (4/3) π R = L , so R = ⎜
⎟
⎝ 4π ⎠
3
4π R 2 4π ⎛ 3 ⎞
The ratio of the areas is
=
=
⎜
⎟
6 ⎝ 4π ⎠
Ac
6 L2
then
As
1/4
3
L.
2/3
= 0.806 . The temperature of the cube is,
⎛A ⎞
1/ 4
Tc = ⎜ s ⎟ Ts = ( 0.806 ) ( 773 K ) = 732 K
⎜A ⎟
⎝ c⎠
Chapter 13 Problems
715
41. REASONING The heat lost per second due to conduction through the glass is given by
Equation 13.1 as Q/t = (kAΔT)/L. In this expression, we have no information for the
thermal conductivity k, the cross-sectional area A, or the length L. Nevertheless, we can
apply the equation to the initial situation and again to the situation where the outside
temperature has fallen. This will allow us to eliminate the unknown variables from the
calculation.
SOLUTION Applying Equation 13.1 to the initial situation and to the situation after the
outside temperature has fallen, we obtain
(
kA TIn − TOut, initial
⎛Q⎞
=
⎜ ⎟
L
⎝ t ⎠ Initial
)
(
kA TIn − TOut, colder
⎛Q⎞
=
⎜ ⎟
L
⎝ t ⎠Colder
and
)
Dividing these two equations to eliminate the common variables gives
( Q / t )Colder
( Q / t )Initial
(
)
(
)
kA TIn − TOut, colder
=
TIn − TOut, colder
L
=
TIn − TOut, initial
kA TIn − TOut, initial
L
Remembering that twice as much heat is lost per second when the outside is colder, we find
2 ( Q / t )Initial
( Q / t )Initial
=2=
TIn − TOut, colder
TIn − TOut, initial
Solving for the colder outside temperature gives
TOut, colder = 2TOut, initial − TIn = 2 ( 5.0 °C ) − ( 25 °C ) = −15 °C
42. REASONING The flow of heat from the heating elements through the pot bottoms and into
the boiling water occurs because the temperature T of each burner is greater than the
temperature T0 = 100.0 °C of the boiling water. The temperature difference ΔT = T – T0
Q kA ΔT
=
drives heat flow at a rate that is given by
(Equation 13.1), where the bottom of
t
L
a pot has a thermal conductivity k, a cross-sectional area A, and a thickness L. The thermal
conductivities of copper and aluminum (kCu, kAl) are different (see Table 13.1), but the two
pot bottoms are identical in every other respect. Further, because both pots are boiling away
Q
through both bottoms. We
at the same rate, the flow of heat must occur at the same rate
t
will use Equation 13.1 and the temperature TAl of the heating element under the aluminum-
716 THE TRANSFER OF HEAT
bottomed pot to determine the temperature TCu of the heating element under the copperbottomed pot.
Q
of heat flow, as well as equal crosst
Q kA ΔT
(Equation 13.1) yields
sectional areas A and thicknesses L. Therefore, =
t
L
SOLUTION Both pot bottoms have identical rates
Q kCu A ( ΔT )Cu kAl A ( ΔT )Al
=
=
t
L
L
kCu ( ΔT )Cu = kAl ( ΔT )Al
or
(1)
Solving Equation (1) for ( ΔT )Cu yields
( ΔT )Cu =
kAl ( ΔT )Al
kCu
(2)
Substituting ( ΔT )Cu = TCu − T0 into Equation (2) and solving for TCu, we obtain
TCu − T0 =
kAl ( ΔT )Al
kCu
or
TCu =
kAl ( ΔT )Al
+ T0
kCu
Therefore, the temperature of the heating element underneath the copper-bottomed pot is
TCu
(
)(
(
)
⎡ 240 J/ s ⋅ m ⋅ CD ⎤ 155.0 DC − 100.0 DC
⎦
=⎣
+ 100.0 DC = 134 DC
D
390 J/ s ⋅ m ⋅ C
)
43. SSM WWW REASONING Heat flows along the rods via conduction, so that Equation
( k AΔT ) t , where Q is the amount of heat that flows in a time t, k is the
13.1 applies: Q =
L
thermal conductivity of the material from which a rod is made, A is the cross-sectional area
of the rod, and ΔT is the difference in temperature between the ends of a rod. In arrangement
a, this expression applies to each rod and ΔT has the same value of ΔT = TW − TC . The total
heat Q ′ is the sum of the heats through each rod. In arrangement b, the situation is more
complicated. We will use the fact that the same heat flows through each rod to determine the
temperature at the interface between the rods and then use this temperature to determine ΔT
and the heat flow through either rod.
Chapter 13 Problems
717
SOLUTION For arrangement a, we apply Equation 13.1 to each rod and obtain for the total
heat that
k A ( TW − TC ) t k2 A ( TW − TC ) t ( k1 + k2 ) A ( TW − TC ) t
Q′ = Q1 + Q2 = 1
+
=
(1)
L
L
L
For arrangement b, we use T to denote the temperature at the interface between the rods and
note that the same heat flows through each rod. Thus, using Equation 13.1 to express the
heat flowing in each rod, we have
k1 A ( TW − T ) t
L
=
k2 A ( T − TC ) t
L
Heat flowing
through rod 1
or
k1 ( TW − T ) = k2 ( T − TC )
Heat flowing
through rod 2
Solving this expression for the temperature T gives
T=
k1TW + k2TC
k1 + k2
(2)
Applying Equation 13.1 to either rod in arrangement b and using Equation (2) for the
interface temperature, we can determine the heat Q that is flowing. Choosing rod 2, we find
that
⎛ kT +k T
⎞
k2 A ⎜⎜ 1 W 2 C − TC ⎟⎟ t
k A ( T − TC ) t
⎝ k1 + k2
⎠
=
Q= 2
L
L
⎛ k T −k T ⎞
k2 A ⎜⎜ 1 W 1 C ⎟⎟ t
⎝ k1 + k2 ⎠ = k2 Ak1 ( TW − TC ) t
=
L
L ( k1 + k2 )
Using Equations (1) and (3), we obtain for the desired ratio that
Q′
=
Q
( k1 + k2 ) A ( TW − TC ) t
L
k2 Ak1 ( TW − TC ) t
L ( k1 + k2 )
( k1 + k2 ) A ( TW − TC ) t L ( k1 + k2 ) ( k1 + k2 )2
=
=
k2 k1
L k2 A k1 ( TW − TC ) t
Using the fact that k2 = 2k1 , we obtain
(3)
718 THE TRANSFER OF HEAT
( k + 2k1 ) = 4.5
Q′ ( k1 + k2 )
=
= 1
Q
k2 k1
2k1 k1
2
2
44. REASONING The drawing shows a crosssectional view of the small sphere inside the larger
spherical asbestos shell. The small sphere produces
a net radiant energy, because its temperature
(800.0 °C) is greater than that of its environment
(600.0 °C). This energy is then conducted through
the thin asbestos shell (thickness = L). By setting
the net radiant energy produced by the small
sphere equal to the energy conducted through the
asbestos shell, we will be able to obtain the
temperature T2 of the outer surface of the shell.
T2
600.0 °C
L
r2
r1
800.0 °C
SOLUTION The heat Q conducted during a time through the thin asbestos shell is given
A ΔT ) t
(k
by Equation 13.1 as Q = asbestos 2
, where kasbestos is the thermal conductivity of
L
asbestos (see Table 13.1), A2 is the surface area of the spherical shell A2 = 4π r22 , ΔT is the
(
)
temperature difference between the inner and outer surfaces of the shell
(ΔT = 600.0 °C − T2), and L is the thickness of the shell. Solving this equation for the T2
yields
QL
T2 = 600.0 °C −
kasbestos 4π r22 t
(
)
The heat Q is produced by the net radiant energy generated by the small sphere inside the
asbestos shell. According to Equation 13.3, the net radiant energy is
Q = Pnet t = eσ A1 T 4 − T04 t , where e is the emissivity, σ is the Stefan-Boltzmann constant,
(
)
A1 is the surface area of the sphere
( A1 = 4π r12 ) , T is the temperature of the sphere
(T = 800.0 °C = 1073.2 K) and T0 is the temperature of the environment that surrounds the
sphere (T0 = 600.0 °C = 873.2 K). Substituting this expression for Q into the expression
above for T2, and algebraically eliminating the time t and the 4π factors, gives
Chapter 13 Problems
(
719
)
⎡ eσ T 4 − T 4 ⎤ L
0 ⎦
T2 = 600.0 °C − ⎣
2
⎛ r2 ⎞
kasbestos ⎜ ⎟
⎝ r1 ⎠
⎧
J
4
4 ⎫
⎛
⎞⎡
−8
1073.2 K ) − (873.2 K ) ⎤ ⎬ 1.0 × 10−2 m
⎨( 0.90 ) ⎜ 5.67 × 10
2
4 ⎟ ⎣(
⎦⎭
s⋅m ⋅K ⎠
⎝
= 600.0 °C − ⎩
J
2
⎡
⎤
⎢⎣0.090 s ⋅ m ⋅ C° ⎥⎦ (10.0 )
(
= 558 °C
)