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JEE Main 2014 (12 Apr, Shift 1) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2014 (12 Apr, Shift 1)

Q72JEE Main 2014 · 12 APR · SHIFT 1Medium

Heat Engines, Refrigerators & Second Law

A Carnot engine absorbs 1000 J of heat energy from a reservoir at 127oC and rejects 600 J of heat energy during each cycle. The efficiency of engine and temperature of sink will be:

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Answer: (b)

  1. η=1−Q2Q1=1−6001000=40%\eta = 1 - \frac{Q_2}{Q_1} = 1 - \frac{600}{1000} = 40\%
  2. T2T1=Q2Q1\frac{T_2}{T_1} = \frac{Q_2}{Q_1}: T2=0.6×400=240 KT_2 = 0.6 \times 400 = 240\text{ K}
  3. T2=240−273=−33∘CT_2 = 240 - 273 = -33^\circ\text{C}
  4. So the answer is (b) 40% and −33∘C-33^\circ\text{C}.
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