Physics Solution Manual for 1100 and 2101

T2 6000 c l r2 r1 8000 c solution the heat q conducted

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Unformatted text preview: noting that P2 = P1, gives T2 = T1 4 4 P2 e σ A2 P 1 e σ A1 =4 A1 A2 P 1 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 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 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 = = Poriginal eσ T 4 A1 11 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 ∆T ) (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, ∆T 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 ( kArod ∆T ) = 390 J ( s ⋅ m ⋅ Co ) (1.2 ×10−4 m 2 ) ( 24 oC − 0 oC ) = 11 W = L 0.10 m 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 ⋅ m 2 ⋅ 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 ∆T ) 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, 79 J / s ⋅ m ⋅ C ° π 5.0 × 10 –3 m ( k A ∆T ) t Q= = L 1.2 m b gc 36. REASONING AND SOLUTION refrigerator walls is 502 b= hb ° C – 26 ° C g5.0 sg 2 12 J The rate at which energy is gained through the ( ) 2 Q kA ∆T [ 0.030 J/(s ⋅ m ⋅ C°) ] 5.3 m ( 25 ...
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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.

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