14_InstSolManual_PDF_Part9

14_InstSolManual_PDF_Part9 - Temperature and Heat 14-9...

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Solve: (a) (b) Heat is radiated away into space at the same rate as heat reaches the surface. 14.52. Set Up and Solve: (a) (b) The power input must be 196 W, to replace the heat conducted through the walls. 14.53. Set Up: Apply and solve for k . Solve: Reflect: This is a small value; skin is a poor conductor of heat. But the thickness of the skin is small, so the rate of heat conduction through the skin is not small. 14.54. Set Up: For water, For steel, Solve: 14.55. Set Up: The heat current is the same through the wood as through the Styrofoam™. Solve: (a) and gives and (b) Or, which checks. Reflect: k is much smaller for the Styrofoam™ so the temperature gradient across it is much larger than across the wood. 14.56. Set Up: For copper, The temperature gradient is Solve: (a) (b) (c) The temperature gradient is the same for all sections of the rod, so and T 5 73.4°C. T H 2 T 5 26.6 C° T H 2 T 12.0 cm 5 2.22 C° / cm. Q t 5 kA 1 T H 2 T C 2 L 5 1 385 W / m # K 21 1.25 3 10 2 4 m 2 21 100 C° 2 0.450 m 5 10.7 W. T H 2 T C L 5 100.0 C° 45.0 cm 5 2.22 C° / cm T H 2 T C L . k 5 385 W / m # K. 1 Q tA 2 s 5 1 0.010 W / m # K 21 19.0°C 2 3 2 5.8°C 42 0.022 m 5 11 W / m 2 , 1 Q tA 2 w 5 k 1 T H 2 T C 2 L 5 1 0.080 W / m # K 21 2 5.8°C 1 10.0°C 2 0.030 m 5 11 W / m 2 T 5 26.7°C 2 8.65°C 2 3.125
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