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Heat supplied to a Carnot egine is `453.6 kcal`. How much useful work can be done by the engine that works between `10^(@)C`and `100^(@)C`? |
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Answer» `T_(2) = 100 +273 = 373 K , T_(1) = 10 + 273 = 283K,` `q_(2) = 453.6 xx 4.184 = 1897.86 kJ` We know that `w = q_(2).(T_(2)-T_(1))/(T_(2))` `= 1897.86 xx ((373-283))/(373)` `= (1897.86 xx 90)/(373) = 457.92 kJ` |
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