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The efficiency of an ideal heat engine working between the freezing point and boiling point of water is |
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Answer» `26.8%` FREEZINGPOINT of water `= 0^(@)C = 273 K` Boiling POINT of water `= 100^(C) = (100 + 273)K = 373 K` `T_(2)` : Sink temperature = 273 K `T_(1)` : Source temperature = 373 K `% eta = (1-(T_(2))/(T_(1))) xx 100 = (1-(273)/(373)) xx 100 = 26.8%` |
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