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InterviewSolution
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A certain engine which operates in a Carnot cycle absorbe `3.0 kJ` at `400^(@)C` in a cycle. If it rejects heat at `100^(@)C`, how much work is done on the engine per cycle and how much heat is evolved at `100^(@)C` in each cycle? |
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Answer» Given, Heat absorbs `(q_(2)) = 3.0 kJ` `T_(2) = 400 + 273.15 K = 673.15 K` `T_(1) = 100 +273.15 K = 373.15 K` Efficiency of the Carnot cycle is given by `eta = (T_(2)-T_(1))/(T_(2)) = (q_(2)+q_(1))/(q_(2))` Thus, `(-T_(1))/(T_(2)) = (q_(1))/(q_(2))` Hence, `q_(1) = ((-T_(1))/(T_(2))) (q_(2))` Thus, heat evolved `q_(1) =- ((373.15K)/(673.15K)) (3.0kJ) =- 1.662 kJ` The work done on the engine is `w =- (q_(2)+q_(1)) = - (3.0 +1.662) kJ =- 4.662kJ` |
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