593_Physics ProblemsTechnical Physics

# 593_Physics ProblemsTechnical Physics - Chapter 20 a fb 595...

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Chapter 20 595 P20.62 (a) Fv == 500 400 2000 .. N m s W a f bg (b) Energy received by each object is 1 000 10 10 2 389 4 W s J c a l af . The specific heat of iron is 0107 . c a l g C ⋅° , so the heat capacity of each object is 500 10 0107 5350 3 . ×× = ° cal C. T = ° 2389 447 cal 535.0 cal C C . P20.63 The power incident on the solar collector is P i IA = 600 0 300 170 2 Wm m W 2 ej π For a 40.0% reflector, the collected power is c = 67 9 . W. The total energy required to increase the temperature of the water to the boiling point and to evaporate it is Qc mTm L V =+ : Q =⋅ ° ° + × = × 0 500 80 0 2 26 10 1 30 10 66 . . kg 4186 Jkg C C Jkg J . The time interval required is
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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