Two ideal Carnot engines operate in cascade (all heat given up by one… | JEE Main 2020 (7 Jan, Shift 2)

Q54JEE Main 2020 · 7 JAN · SHIFT 2Difficult

Heat Engines, Refrigerators & Second Law

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperature, T1 and T2. The temperature of the hot reservoir of the first engine is T1 and the temperature of the cold reservoir of the second engine is T2. T is temperature of the sink of first engine which is also the source for the second engine. How is T related to T1 and T2, if both engines perform equal amount of work?

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Answer: (c)

First engine: $W_1 = Q_1\left(1 - \frac{T}{T_1}\right)$ , rejects $Q = Q_1\frac{T}{T_1}$ .
Second engine: $W_2 = Q\left(1 - \frac{T_2}{T}\right) = \frac{Q_1(T - T_2)}{T_1}$
$W_1 = W_2 \Rightarrow T_1 - T = T - T_2 \Rightarrow T = \frac{T_1 + T_2}{2}$
So the answer is (c) $T = \frac{T_1 + T_2}{2}$ .

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Heat Engines, Refrigerators & Second Law question from JEE Main 2020 (7 Jan, Shift 2)

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