Two Carnot engines A and B are operated in series. Engine A receives… | JEE Main 2018 (15 Apr, Shift 2)

Q62JEE Main 2018 · 15 APR · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

Two Carnot engines A and B are operated in series. Engine A receives heat from a reservoir at 600 K and rejects heat to a reservoir at temperature T. Engine B receives heat rejected by engine A and in turn rejects it to a reservoir at 100 K. If the efficiencies of the two engines A and B are represented by $\eta_A$ and $\eta_B,$ respectively, then what is the value of $\frac{\eta_B}{\eta_A}$ ?

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Answer: (a)

Assuming equal work outputs (the original condition): $T = \frac{600 + 100}{2} = 350\text{ K}$ .
$\eta_A = 1 - \frac{350}{600} = \frac{5}{12}$
$\eta_B = 1 - \frac{100}{350} = \frac{5}{7}$
$\frac{\eta_B}{\eta_A} = \frac{5/7}{5/12} = \frac{12}{7}$ . So the answer is (a) $\frac{12}{7}$ .

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