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587_Physics ProblemsTechnical Physics

# 587_Physics ProblemsTechnical Physics - Chapter 20 Section...

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Chapter 20 589 Section 20.7 Energy Transfer Mechanisms P20.41 P = kA T L k L AT == ° ° 10 0 120 150 222 10 2 . .. . W 0.040 0 m mC Wm C 2 bg a f P20.42 ⋅° ° × = kA T L 0800 300 250 600 10 100 10 100 3 4 . . m C m W k W 2 ej af P20.43 In the steady state condition, PP Au Ag = so that kA T x T x Au Au Au Ag Ag Ag F H G I K J = F H G I K J In this case AA Au Ag = ∆∆ xx TT Au Ag Au = =− 80 0 . and Ag 30 0 . FIG. P20.43 where T is the temperature of the junction. Therefore, kT k T Au Ag 80 0 30 0 −= And T 51 2 .C P20.44 ° ×⋅ ° + × ° = −− L k i i i 600 500 2 4 00 10 0 800 5 00 10 0 023 4 134 33 . . . m W m C m W m C kW 2 *P20.45 We suppose that the area of the transistor is so small that energy flow by heat from the transistor directly to the air is negligible compared to energy conduction through the mica. = =+ = °+ × ⋅° × kA L L kA hc a f 35 0 1 50 0 085 2 10 00753 825 62510 67 9 3 6 . . . C m m C 2 P20.46 From Table 20.4,
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