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JEE Main 2022 (27 Jul, Shift 2) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2022 (27 Jul, Shift 2)

Q33JEE Main 2022 · 27 JUL · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

In 11st case, Carnot engine operates between temperatures 300 K300\text{ K} and 100 K100\text{ K}. In 22nd case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in 22nd case) will be:

Figure

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Answer: (a)

  1. 1st case: η=1−100300=23\eta = 1 - \frac{100}{300} = \frac{2}{3}.
  2. 2nd case: Q1=Q_1 = heat from 300 K, Q3=Q_3 = heat to 100 K; for reversible engines Q3Q1=Q3Q2⋅Q2Q1=100200⋅200300=13\frac{Q_3}{Q_1} = \frac{Q_3}{Q_2}\cdot\frac{Q_2}{Q_1} = \frac{100}{200}\cdot\frac{200}{300} = \frac{1}{3}.
  3. ηcomb=1−Q3Q1=23\eta_{\text{comb}} = 1 - \frac{Q_3}{Q_1} = \frac{2}{3}
  4. So the answer is (a) same as the 1st case.
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