589_Physics ProblemsTechnical Physics

589_Physics ProblemsTechnical Physics - Chapter 20 591...

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Chapter 20 591 Additional Problems P20.52 77.3 K = –195.8°C is the boiling point of nitrogen. It gains no heat to warm as a liquid, but gains heat to vaporize: Qm L v == × 0 100 2 01 10 2 01 10 54 .. . kg J kg J bg ej . The water first loses heat by cooling. Before it starts to freeze, it can lose cT ° ° = × 0 200 4186 5 00 4 19 10 3 . kg J kg C C J b g af . The remaining 2 01 10 4 19 10 1 59 10 43 4 . ×− × = × J J that is removed from the water can freeze a mass x of water: L x x f = ×= × 1 59 10 3 33 10 00477 477 45 J J k g kg g of water can be frozen P20.53 The increase in internal energy required to melt 1.00 kg of snow is E int kg J kg J = × 1 00 3 33 10 3 33 10 55 . The force of friction is fnm g = = µµ 0 200 75 0 9 80 147 . kg m s N 2 According to the problem statement, the loss of mechanical energy of the skier is assumed to be equal to the increase in internal energy of the snow. This increase in internal energy is ∆∆ Ef r r int N J = × 147 3 33 10 5 . and r 227 10 3 . m . P20.54 (a) The energy thus far gained by the copper equals the energy loss by the silver. Your down
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