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A perfect carnot engine utilizes an ideal gas the source temperature is500K and sink temperature is 375K. If the engine takes 600k cal per cyclefrom the source, calculate(i) The efficiency of engine(ii) Work done per cycle(iii) Heat rejected to sink per cycle.
Dear Vaibhav T1 = 500 K T2 = 375 K Q1 = heat absorbed = 600 k cal i) efficiency (n) = 1 -(T2/T1) = 125/500 = 0.25 = 25% ii) n = W/Q1 W = n* Q1 = 0.25 * 600 = 150 k cal iii)- W = Q1 - Q2 Q2 = Q1 - W Q2 = 600 - 150 = 450 k cal Regards Arun (askIITians forum expert)
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