Question:
Two Carnot engines A and B are operated in series. The first one, A, receives heat at \(T_1\) (= 600 K) and rejects to a reservoir at temperature \(T_2\). The second engine B receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at \(T_3\) (= 400 K). Calculate the temperature \(T_2\) if the work outputs of the two engines are equal :
Two Carnot engines A and B are operated in series. The first one, A, receives heat at \(T_1\) (= 600 K) and rejects to a reservoir at temperature \(T_2\). The second engine B receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at \(T_3\) (= 400 K). Calculate the temperature \(T_2\) if the work outputs of the two engines are equal :
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For Carnot engines Q is proportional to T, so equal work means T1 - T2 = T2 - T3.
Updated On: Nov 5, 2026
- 600 K
- 400 K
- 300 K
- 500 K
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The Correct Option is D
Solution and Explanation
For a Carnot engine \(\frac{Q_{\text{in}}}{T_{\text{hot}}} = \frac{Q_{\text{out}}}{T_{\text{cold}}}\), so heat exchanged is proportional to the temperature of the reservoir.
Let engine A take \(Q_1\) at \(T_1\), reject \(Q_2\) at \(T_2\); engine B takes \(Q_2\) and rejects \(Q_3\) at \(T_3\).
\(W_A = Q_1 - Q_2\), \(W_B = Q_2 - Q_3\). Equal work: \(Q_1 - Q_2 = Q_2 - Q_3\).
Since \(Q \propto T\): \(T_1 - T_2 = T_2 - T_3 \Rightarrow T_2 = \frac{T_1 + T_3}{2} = \frac{600 + 400}{2} = 500\) K.
Answer: (4) 500 K
Let engine A take \(Q_1\) at \(T_1\), reject \(Q_2\) at \(T_2\); engine B takes \(Q_2\) and rejects \(Q_3\) at \(T_3\).
\(W_A = Q_1 - Q_2\), \(W_B = Q_2 - Q_3\). Equal work: \(Q_1 - Q_2 = Q_2 - Q_3\).
Since \(Q \propto T\): \(T_1 - T_2 = T_2 - T_3 \Rightarrow T_2 = \frac{T_1 + T_3}{2} = \frac{600 + 400}{2} = 500\) K.
Answer: (4) 500 K
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