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593_Physics ProblemsTechnical Physics

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

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Chapter 20 595 P20.62 (a) Fv = = 50 0 40 0 2 000 . . N m s W a fb g (b) Energy received by each object is 1 000 10 10 2 389 4 W s J cal b ga f = = . The specific heat of iron is 0 107 . cal g C ⋅° , so the heat capacity of each object is 5 00 10 0 107 535 0 3 . . . × × = ° cal C. T = ° = ° 2 389 4 47 cal 535.0 cal C C . P20.63 The power incident on the solar collector is P i IA = = = 600 0 300 170 2 W m m W 2 e j a f π . . For a 40.0% reflector, the collected power is P c = 67 9 . W. The total energy required to increase the temperature of the water to the boiling point and to evaporate it is Q cm T mL V = + : Q = ⋅° ° + × = × 0 500 80 0 2 26 10 1 30 10 6 6 . . . . kg 4186 J kg C C J kg J b ga f . The time interval required is
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