526_PartUniversity Physics Solution

# 526_PartUniversity Physics Solution - 17-12 Chapter 17...

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17-12 Chapter 17 17.67. I DENTIFY and S ET U P : Call the temperature at the interface between the wood and the styrofoam T . The heat current in each material is given by H C ( )/ . H kA T T L = See Figure 17.67 Heat current through the wood: w w 1 w ( ) H k A T T L = Heat current through the styrofoam: s s 2 s ( )/ H k A T T L = Figure 17.67 In steady-state heat does not accumulate in either material. The same heat has to pass through both materials in succession, so w s . H H = E XECUTE : (a) This implies w 1 w s 2 s ( )/ ( )/ k A T T L k A T T L = w s 1 s w 2 ( ) ( ) k L T T k L T T = w s 1 s w 2 w s s w 0.0176 W C/K 00057 W C/K 5.8 C 0.00206 W/K k LT k L T T k L k L + ⋅° + ⋅° = = = − ° + E VALUATE : The temperature at the junction is much closer in value to 1 T than to 2 . T The styrofoam has a very large k , so a larger temperature gradient is required for than for wood to establish the same heat current. (b) I DENTIFY and S ET U P : Heat flow per square meter is H C . H T T k A L = We can calculate this either for the wood or for the styrofoam; the results must be the same. E XECUTE : wood 2 w 1 w w ( 5.8 C ( 10.0 C)) (0.080 W/m K) 11 W/m . 0.030 m H T T k A L ° − − ° = = = styrofoam 2 s 2 s s (19.0 C ( 5.8 C)) (0.010 W/m K) 11 W/m . A 0.022 m H T T k L ° − − ° = = = E VALUATE : H must be the same for both materials and our numerical results show this. Both materials are good insulators and the heat flow is very small. 17.68. I DENTIFY : H C ( ) Q kA T T t L = S ET U P : H C 175 C 35 C. T T = ° ° 1 K 1 C , = ° so there is no need to convert the temperatures to kelvins. E XECUTE : (a) 2 2 (0.040 W/m K)(1.40 m )(175 C 35 C) 196 W. 4.0 10 m Q t = = × ° ° (b) The power input must be 196 W, to replace the heat conducted through the walls. E VALUATE : The heat current is small because k is small for fiberglass. 17.69. I DENTIFY : Apply Eq.(17.23). . Q Ht = S ET U P : 1 Btu 1055 J = E XECUTE : The energy that flows in time t is 2 5 2 (125 ft )(34 F ) (5.0 h) 708 Btu 7.5 10 J. (30 ft F h/Btu) A T Q Ht t R Δ ° = = = = = × °⋅ E VALUATE : With the given units of R , we can use A in 2 ft , T Δ in F ° and t in h, and the calculation then gives Q in Btu. 17.70. I DENTIFY : . Q kA T t L Δ = / Q t is the same for both sections of the rod. S ET U P : For copper, c 385 W/m K. k = For steel, s 50.2 W/m K. k = E XECUTE : (a) For the copper section, 4 2 (385 W/m K)(4.00 10 m )(100 C 65.0 C) 5.39 J/s. 1.00 m Q t × = = ° ° (b) For the steel section, 4 2 (50.2 W/m K)(4.00 10 m )(65.0 C 0 C) 0.242 m. ( / ) 5.39 J/s kA T L Q t Δ × = = = ° ° E VALUATE : The thermal conductivity for steel is much less than that for copper, so for the same T Δ and A a smaller L for steel would be needed for the same heat current as in copper.

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Temperature and Heat 17-13 17.71. I DENTIFY and S ET U P : The heat conducted through the bottom of the pot goes into the water at 100°C to convert it to steam at 100°C. We can calculate the amount of heat flow from the mass of material that changes phase. Then use Eq.(17.21) to calculate H , T the temperature of the lower surface of the pan.
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