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JEE Main 2024 · Physics · Thermodynamics and Kinetic Theory of Gases · medium

The efficiency of a Carnot engine is 40%. When the temperature of its sink is lowered by 60 K, keeping the source fixed, the efficiency becomes 52%.…

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The efficiency of a Carnot engine is 40%40\%. When the temperature of its sink is lowered by 60 K60\ \mathrm{K}, keeping the source fixed, the efficiency becomes 52%52\%. The temperature of the source is
  1. A. 500 K500\ \mathrm{K}
  2. B. 600 K600\ \mathrm{K}
  3. C. 400 K400\ \mathrm{K}
  4. D. 750 K750\ \mathrm{K}
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Answer: (A) 500 K500\ \mathrm{K}

Let the source be at T1T_1 and the sink at T2T_2. From η=1−T2T1\eta=1-\dfrac{T_2}{T_1}: 0.40.4 gives T2=0.6T1T_2=0.6T_1. After lowering the sink: 1−T2−60T1=0.521-\dfrac{T_2-60}{T_1}=0.52 gives T2−60=0.48T1T_2-60=0.48T_1. Substituting, 0.6T1−60=0.48T10.6T_1-60=0.48T_1, so 0.12T1=600.12T_1=60 and T1=500 KT_1=500\ \mathrm{K} (sink at 300 K300\ \mathrm{K}). Check: 1−240/500=0.521-240/500=0.52.
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