526_PartUniversity Physics Solution

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

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17-12 Chapter 17 17.67. IDENTIFY and SET UP: Call the temperature at the interface between the wood and the styrofoam T . The heat current in each material is given by HC () / . H kA T T L =− See Figure 17.67 Heat current through the wood: ww 1 w H kAT TL Heat current through the styrofoam: ss2 s / H kAT 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 ws . H H = EXECUTE: (a) This implies w1 w s 2 s / / kAT T L kAT T L −=− 1 sw 2 kLT T kL T T −= ws1 sw2 sw 0.0176 W C/K 00057 W C/K 5.8 C 0.00206 W/K kLT kLT T kL kL +− ° + ⋅ ° == = ° + EVALUATE: 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) IDENTIFY and SET UP: Heat flow per square meter is . HT T k AL ⎛⎞ = ⎜⎟ ⎝⎠ We can calculate this either for the wood or for the styrofoam; the results must be the same. EXECUTE: wood 2 w w ( 5.8 C ( 10.0 C)) (0.080 W/m K) 11 W/m . 0.030 m T k −− ° ° = styrofoam 2 s2 s s (19.0 C ( 5.8 C)) (0.010 W/m K) 11 W/m . A 0.022 m T k L −° ° = EVALUATE: 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. IDENTIFY: Qk A TT tL = SET UP: 175 C 35 C. °° 1 K 1 C , = ° so there is no need to convert the temperatures to kelvins. EXECUTE: (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. EVALUATE: The heat current is small because k is small for fiberglass. 17.69. IDENTIFY: Apply Eq.(17.23). . QH t = SET UP: 1 Btu 1055 J = EXECUTE: 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) AT t t R Δ ° = = ⋅° EVALUATE: 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. IDENTIFY: . A T Δ = / Qt is the same for both sections of the rod. SET UP: For copper, c 385 W/m K. k =⋅ For steel, s 50.2 W/m K. k = EXECUTE: (a) For the copper section, 42 (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, (50.2 W/m K)(4.00 10 m )(65.0 C 0 C) 0.242 m. (/ ) 5 . 3 9 J / s kA T L Δ⋅ × = EVALUATE: 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. IDENTIFY and SET UP: 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. EXECUTE: 35 v (0.390 kg)(2256 10 J/kg) 8.798 10 J Qm L == × = × 53 / 8.798 10 J/180 s 4.888 10 J/s HQ t × = × Then HC () / H kA T T L =− says that 33 2 (4.888 10 J/s)(8.50 10 m) 5.52 C (50.2 W/m K)(0.150 m ) HL TT kA × × = ° 5.52 C 100 C 5.52 C 105.5 C = + °= ° + ° EVALUATE: The larger is the larger H is and the faster the water boils.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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

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