Unformatted text preview: ctivity, A is the area, ∆T 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 endtoend 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 ( ∆T ) t
La Qb = and kAb ( ∆T ) t
Lb Note that the thermal conductivity k, the temperature difference ∆T, and the time t are the
same in both arrangements. Dividing Qa by Qb gives kAa ( ∆T ) t
Qa
La
AL
=
= ab
Qb kAb ( ∆T ) 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 )( 2 Lb ) = 1
4 ______________________________________________________________________________
5. SSM REASONING The heat transferred in a time t is given by Equation 13.1,
Q = ( k A ∆T ) 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 ∆T
Lal
This expression can be solved for Lst . = kst A ∆T
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 ∆T ) t (Equation 13.1), where k is the thermal
crosssectional area A is Q =
L
conductivity, and ∆T is the temperature difference.
SOLUTION The crosssectional area and length of each block are: AA = 2 L2 and
0
LA = 3 L0 , AB = 3L2 and LB = 2 L0 , AC = 6 L2 and LC = L0 . The heat conducted through each
0
0 block is:
Case A QA =
= 2 L2
AA
k ∆T t = 0 k ∆T t =
LA
3L0
2
3 ( 2 L0 ) k ∆T t
3 ( 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 ∆T t = 0 k ∆T t =
LB
2 L0
3
2 3
( 2 L0 ) k ∆T t ( 0.30 m ) 250 J/ ( s ⋅ m ⋅ C°) ( 35 °C − 19 °C )( 5.0 s ) = 9.0 × 103 J Case C QC = AC
6 L2
k ∆T t = 0 k ∆T t = ( 6 L0 ) k ∆T 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 crosssectional 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 crosssectional 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
crosssectional 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
crosssectional 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
2
A = Aout − Ain = π rout − π rin (1) The heat Q that flows along the pipe in a time t = 15 min is related to the crosssectional
( kA ∆T ) 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 ∆T 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: ∆T = 25 o C − 4.0 o C = 21 C o . 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
2
π rin = π rout − A or 2
2
rin = rout − A π or 2
rin = rout − A π (2) 692 THE TRANSFER OF HEAT An expression for the crosssectional area A of the pipe may be obtained by solving
( kA ∆T ) t (Equation 13.1):
Q=
L
QL
(3)
A=
( k ∆T ) t
Substituting Equation (3) into Equation (2) yields
2
rin = rout − QL
π ( k ∆T ) 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 ⋅ Co 25 oC − 4.0 oC 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 ∆T ) 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 ∆T 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
∆T = ( 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 temp...
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 Spring '13
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 Physics, The Lottery

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