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Unformatted text preview: C° ) Using this value for θ in Equation (1) gives
T= W
85.0 N
=
= 996 N
2sin θ 2 sin 2.446° CHAPTER 13 THE TRANSFER OF HEAT ANSWERS TO FOCUS ON CONCEPTS QUESTIONS ( k A ∆T ) t , 1. (d) The heat conducted during a time t through a bar is given by Q = 2. (b) This arrangement conducts more heat for two reasons. First, the temperature difference
∆T between the ends of each bar is greater in A than in B. Second, the crosssectional area
available for heat conduction is twice as large in A as in B. A greater crosssectional area
means more heat is conducted, everything else remaining the same. 3. 4. (e) This arrangement conducts more heat for two reasons. First, the crosssectional area A
available for heat flow is twice as large in B than in A. Twice the crosssectional area means
twice the heat that is conducted. Second, the length of the bars in B is onehalf 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 ∆T ) t . The heat Q,
(e) The heat conducted through a material is given by Q =
L
crosssectional area A, thickness L, and time t are the same for the three materials. Thus, the
product k ∆T must also be the same for each. Since the temperature difference ∆T 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 = where k is
L
the thermal conductivity, and A and L are the crosssectional area and length of the bar. ( k A ∆T ) t (Equation 13.1). The heat
L
Q, crosssectional area A, length L, and time t are the same for the two smaller bars. Thus, the
product k ∆T must also be the same for each. Since the thermal conductivity k1 is greater than
k2, the temperature difference ∆T across the left bar is smaller than that across the right bar.
Thus, the temperature where the two bars are joined together (400 °C − ∆T) 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
4 4 that it depends on the product of T and the surface area A of the cube. The product T A is
equal to 1944 T04 L2 , 1536 T04 L2 , and 864 T04 L2 for B, A, and C, respectively.
0
0
0
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 blackpainted 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
4
objects, the product eT 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 ∆T ) 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 ∆T 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 =
∆T =
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= ( kA∆T ) t
=
L c b
h g g3600 s I
b F1 h J
GK
H 0.045 J / (s ⋅ m ⋅ C ° ) 1.6 m 2 160 ° C − 50 ° C 6.0 h
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
–7
that 1 J = 2.78 × 10 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 ∆ T [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 Chapter 13 Problems 689 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 A∆T ) t Q= (13.1) L where k is the thermal condu...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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