In a Carnot cycle, heat absorbed at hot reservoir is 1000 J. Efficiency 25%. Heat rejected at cold reservoir:
A1000 J
B500 J
C750 J
D250 J
Answer & Solution
Correct answer: C. 750 J
Work done = η × Q_h = 0.25 × 1000 = 250 J. By 1st law (cyclic): Q_h = W + Q_c → Q_c = Q_h - W = 1000 - 250 = 750 J.
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