One mole of water is cooled from 25.0oC to 0oC and frozen in a refrigerator. All of the heat taken bythe refrigerator, operating at maximum theoretical efficiency (no entropy created) is delivered to asecond mole of water initially at 25.0oC, heating it to 100oC and converting a fraction of it to vapor.What is the entropy change in the first mole of water?
The second mole is confusing me. Can't I just find the Q lost as the water changes from 25 to 0 degrees celsius, integrate Q/T from 0 to 25 (convert these to Kelvin, obviously) then find the Q for the freezing of one mole of water and divide by 273.15K? Second mole shouldn't matter? Or do I, for some reason, have to find the Q lost by the cooling/freezing water, subtract from that the Q needed to heat the water from 25c to 100c, use the remaining to get the entropy gained by vaporization and also calculate the entropy gained by heating and find the negative?
Answer is -28.6 J/K. Thanks!
Submitted December 10, 2015 at 08:07AM by JShrub http://ift.tt/1Rb9TXV HomeworkHelp
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