Question:
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 \(\dfrac{\eta_B}{\eta_A}\) ?
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 \(\dfrac{\eta_B}{\eta_A}\) ?
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The value of T is not stated; the intended condition is that both engines do equal work, which gives T = 350 K.
Updated On: Nov 11, 2026
- \(\frac{12}{7}\)
- \(\frac{7}{12}\)
- \(\frac{12}{5}\)
- \(\frac{5}{12}\)
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The Correct Option is A
Solution and Explanation
T is not given explicitly; the standard intended assumption is that the two engines in series do equal work (\(W_A = W_B\)).
Let \(Q_1\) be the heat taken by A from the 600 K reservoir. For Carnot engines, heat rejected by A is \(Q_2 = Q_1 T/600\), and heat rejected by B is \(Q_3 = Q_2 \cdot 100/T\).
\(W_A = Q_1 - Q_2 = Q_1\dfrac{600 - T}{600}\) and \(W_B = Q_2 - Q_3 = Q_2\dfrac{T-100}{T} = Q_1\dfrac{T - 100}{600}\).
Setting \(W_A = W_B\): \(600 - T = T - 100\), so \(T = 350\) K.
\(\eta_A = 1 - \dfrac{350}{600} = \dfrac{5}{12}\), \(\eta_B = 1 - \dfrac{100}{350} = \dfrac{5}{7}\).
\(\dfrac{\eta_B}{\eta_A} = \dfrac{5/7}{5/12} = \dfrac{12}{7}\).
Answer: (1) \(\frac{12}{7}\)
Let \(Q_1\) be the heat taken by A from the 600 K reservoir. For Carnot engines, heat rejected by A is \(Q_2 = Q_1 T/600\), and heat rejected by B is \(Q_3 = Q_2 \cdot 100/T\).
\(W_A = Q_1 - Q_2 = Q_1\dfrac{600 - T}{600}\) and \(W_B = Q_2 - Q_3 = Q_2\dfrac{T-100}{T} = Q_1\dfrac{T - 100}{600}\).
Setting \(W_A = W_B\): \(600 - T = T - 100\), so \(T = 350\) K.
\(\eta_A = 1 - \dfrac{350}{600} = \dfrac{5}{12}\), \(\eta_B = 1 - \dfrac{100}{350} = \dfrac{5}{7}\).
\(\dfrac{\eta_B}{\eta_A} = \dfrac{5/7}{5/12} = \dfrac{12}{7}\).
Answer: (1) \(\frac{12}{7}\)
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