A boy swimming comes out from a pool covered with a film of water weigh 18g.how much heat must be supplied to evaporate this water at 298K .calculate the internal energy of vaporisation at 100C .the enthalpy change of vapourisation for water ta 373K=40.66 kJ/mol

Asked by  | 5th Oct, 2013, 08:34: AM

Expert Answer:

18 g water = 1 mole water. The temp. is 298K. To raise the temp. to bp (373K), it absorbs 18 x (373 -298) cal (=18 x 4.185 x (373 -298)J) heat and another 40660 j to Liquid --> vapour phase. Sum = [18x4.185{373 - 298} +40660] x (1/1000)kJ.

Answered by  | 5th Oct, 2013, 04:00: PM

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