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JEE Main 2018 (15 Apr, Shift 1) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2018 (15 Apr, Shift 1)

Q63JEE Main 2018 · 15 APR · SHIFT 1Medium

Heat Engines, Refrigerators & Second Law

A Carnot's engine works as a refrigerator between 250250 K and 300300 K. It receives 500500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is:

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Answer: (a)

  1. β=T2T1−T2=250300−250=5\beta = \frac{T_2}{T_1 - T_2} = \frac{250}{300 - 250} = 5
  2. Q2=500 cal=500×4.2=2100Q_2 = 500\text{ cal} = 500 \times 4.2 = 2100 J
  3. W=Q2β=21005=420W = \frac{Q_2}{\beta} = \frac{2100}{5} = 420 J
  4. So the answer is (a) 420420 J.
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