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Unformatted text preview: noting that P2 = P1, gives T2
=
T1 4 4 P2
e Ïƒ A2
P
1
e Ïƒ A1 =4 A1
A2 P
1
e Ïƒ 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 L2
0
28 L2
0 = 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 StefanBoltzmann law: Q = eÏƒ T 4 At . The emitted power is P = Q / t = eÏƒ 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 eÏƒ T 4 A
2
=
(1)
Poriginal eÏƒ 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 crosssectional radius of the original cylinder, with L = 10 r , A1 = (area of ends) + (area of cylinder body)
= 2(Ï€ r 2 ) + (2Ï€ r ) L = 2(Ï€ r 2 ) + (2Ï€ r )(10r ) = 22Ï€ 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(Ï€ r 2 ) + (2Ï€ r ) L = N 2(Ï€ r 2 ) + (2Ï€ r )(10r ) = ( 2 N + 20 ) Ï€ 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 eÏƒ T 4 A
( 2 N + 20 ) Ï€ r 2 = N + 10
2
=
=
Poriginal eÏƒ T 4 A1
11
22Ï€ 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 âˆ†T ) (13.1) L where k is the thermal conductivity of copper (see Table 13.1 in the text), Arod is the
crosssectional area of the rod, âˆ†T is the temperature difference between the wall (24 Â°C)
and the ice (0 Â°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 = eÏƒ Asphere T 4 âˆ’ T04 ) (13.3) where e is the emissivity of the ice, Ïƒ is the StefanBoltzmann 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 sphereâ€™s radius R:
L = 0.25 m âˆ’ 0.15 m = 0.10 m. With this substitution, Equation 13.1 gives the sphereâ€™s
conductive power gain: Pcon ( kArod âˆ†T ) = 390 J ( s â‹… m â‹… Co ) (1.2 Ã—10âˆ’4 m 2 ) ( 24 oC âˆ’ 0 oC ) = 11 W =
L 0.10 m The surface area of a sphere is given by Asphere = 4Ï€R2, so that the net radiant power gain
given by Equation 13.3 becomes ( ) ( Prad = eÏƒ Asphere T 4 âˆ’ T04 = 4Ï€ eÏƒ R 2 T 4 âˆ’ 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 Â°C + 273 = 273 K, and the temperature of the room is T = 24 Â°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 = 4Ï€ eÏƒ R 2 T 4 âˆ’ T04 )
( ) 2
4
4
= 4Ï€ ( 0.90 ) 5.67 Ã— 10âˆ’8 J/ s â‹… m 2 â‹… K 4 ( 0.15 m ) ( 297 K ) âˆ’ ( 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 âˆ†T ) t / L . If we assume that the poker has a circular crosssection, then its
crosssectional area is A = Ï€ r 2 . Table 13.1 gives the thermal conductivity of iron as
79 J / (s â‹… m â‹… C Â° ) .
SOLUTION The amount of heat conducted from one end of the poker to the other in 5.0 s
is, therefore, 79 J / s â‹… m â‹… C Â° Ï€ 5.0 Ã— 10 â€“3 m
( k A âˆ†T ) t
Q=
=
L
1.2 m b gc 36. REASONING AND SOLUTION
refrigerator walls is 502
b=
hb Â° C â€“ 26 Â° C g5.0 sg
2 12 J The rate at which energy is gained through the ( ) 2
Q kA âˆ†T [ 0.030 J/(s â‹… m â‹… CÂ°) ] 5.3 m ( 25 Â...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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