Two Carnot engines A and B operate in series such that engine A… | JEE Main 2021 (27 Jul, Shift 2)

Q49JEE Main 2021 · 27 JUL · SHIFT 2Difficult

Heat Engines, Refrigerators & Second Law

Two Carnot engines A and B operate in series such that engine A absorbs heat at T1 and rejects heat to a sink at temperature T. Engine B absorbs half of the heat rejected by Engine A and rejects heat to the sink at T3. When workdone in both the cases is equal, to value of T is:

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Answer: (d)

Engine A: $W_A = Q_1\left(1 - \frac{T}{T_1}\right)$ , heat rejected $= Q_1\frac{T}{T_1}$ .
Engine B takes half of it: $W_B = \frac{Q_1T}{2T_1}\left(1 - \frac{T_3}{T}\right) = \frac{Q_1(T - T_3)}{2T_1}$
$W_A = W_B \Rightarrow 2(T_1 - T) = T - T_3 \Rightarrow T = \frac{2T_1 + T_3}{3}$
So the answer is (d) $\frac{2}{3}T_1 + \frac{1}{3}T_3$ .

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