577_Physics ProblemsTechnical Physics

577_Physics ProblemsTechnical Physics - Chapter 20 579...

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Chapter 20 579 SOLUTIONS TO PROBLEMS Section 20.1 Heat and Internal Energy P20.1 Taking m = 100 . kg, we have Um g h g == = 1 00 9 80 50 0 490 .. . kg m s m J 2 bg ej a f . But ∆∆ UQ m c T T g = ° = 4186 490 . k g J k gC J b g so T 0117 .C TT T fi =+ = + ° 10 0 0 117 af C P20.2 The container is thermally insulated, so no energy flows by heat: Q = 0 and EQ W W m g h int =+ = input input 02 The work on the falling weights is equal to the work done on the water in the container by the rotating blades. This work results in an increase in internal energy of the water: 2 mgh E m c T int water T mgh mc × ⋅° = ° 2 2 1 50 9 80 3 00 0 200 88 2 0105 water 2 kg m s m kg J kg C J 837 J C C . . . . FIG. P20.2 Section 20.2 Specific Heat and Calorimetry P20.3 Qm c T = silver 123 0525 0234 . . kJ kg C kJ kg C silver silver =⋅ ° c c P20.4 From cT =∆
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