Chapter 20591Additional ProblemsP20.5277.3 K =–195.8°C is the boiling point of nitrogen. It gains no heat to warm as a liquid, but gains heatto vaporize:QmLv==×=×0 1002 01 102 01 1054...kgJ kgJbgej.The water first loses heat by cooling. Before it starts to freeze, it can losecT⋅°°=×∆0 20041865 004 19 103.kgJ kg CCJbgaf.The remaining 2 01 104 19 101 59 10434.×−×=×JJ that is removed from the water can freeze amass xof water:Lxxf=×=×1 59 103 33 100047747745J Jkgkgg of water can be frozenP20.53The increase in internal energy required to melt 1.00 kg of snow is∆EintkgJ kgJ=×1 003 33 103 33 1055.The force of friction isfnmg==µµ0 200 75 09 80147.kgm sN2According to the problem statement, the loss of mechanical energy of the skier is assumed to beequal to the increase in internal energy of the snow. This increase in internal energy is∆∆∆EfrrintN J=×1473 33 105.and∆r227 103. 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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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 .