Thermodynamics

JEE Main level, single correct, medium

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Question 1 of 190

A Carnot engine working between temperatures T1=500 KT_1 = 500\,\text{K} (source) and T2=300 KT_2 = 300\,\text{K} (sink) absorbs 1000 J1000\,\text{J} of heat per cycle. The work done per cycle is:

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Answer: (C) 400 J400\,\text{J}

Efficiency η=1−T2T1=1−300500=0.4\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{500} = 0.4. Work done W=ηQ1=0.4×1000=400 JW = \eta Q_1 = 0.4 \times 1000 = 400\,\text{J}.

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