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JEE Main 2021 (27 Aug, Shift 2) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2021 (27 Aug, Shift 2)

Q46JEE Main 2021 · 27 AUG · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

A heat engine operates between a cold reservoir at temperature T2 = 400 K and a hot reservoir at temperature T1. It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be ______________ K.

Numerical answer type. Work it out, then open the answer.

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Answer: 500

  1. Actual efficiency η=1−240300=0.2\eta = 1 - \frac{240}{300} = 0.2.
  2. It cannot exceed Carnot: 0.2≤1−400T1⇒T1≥4000.80.2 \le 1 - \frac{400}{T_1} \Rightarrow T_1 \ge \frac{400}{0.8}.
  3. Minimum T1=500 KT_1 = 500\text{ K}
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