Question:
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 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?
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 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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Both engines are Carnot, so \(\dfrac{Q}{T}\) is the same at each stage (\(Q_1/T_1=Q/T=Q_2/T_2\)). Equal work means \(Q_1-Q=Q-Q_2\).
Updated On: Oct 31, 2026
- \( T = {{2{T_1}{T_2}} \over {{T_1} + {T_2}}} \)
- \( T = \sqrt {{T_1}{T_2}} \)
- \( T = {{{T_1} + {T_2}} \over 2} \)
- T = 0
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The Correct Option is C
Solution and Explanation
Step 1: Engine 1 takes \(Q_1\) at \(T_1\), rejects \(Q\) at T; engine 2 takes \(Q\) at T and rejects \(Q_2\) at \(T_2\).
Step 2: Equal work: \(Q_1-Q=Q-Q_2\Rightarrow Q=\dfrac{Q_1+Q_2}{2}\).
Step 3: For Carnot engines \(Q\propto T\) at each reservoir: \(\dfrac{Q_1}{T_1}=\dfrac Q T=\dfrac{Q_2}{T_2}\), so \(Q_1,Q_2\) are proportional to \(T_1,T_2\) with the same constant that gives \(Q\propto T\).
Step 4: Hence \(T=\dfrac{T_1+T_2}{2}\).
Answer: (C) \(T=\dfrac{T_1+T_2}{2}\)
Step 2: Equal work: \(Q_1-Q=Q-Q_2\Rightarrow Q=\dfrac{Q_1+Q_2}{2}\).
Step 3: For Carnot engines \(Q\propto T\) at each reservoir: \(\dfrac{Q_1}{T_1}=\dfrac Q T=\dfrac{Q_2}{T_2}\), so \(Q_1,Q_2\) are proportional to \(T_1,T_2\) with the same constant that gives \(Q\propto T\).
Step 4: Hence \(T=\dfrac{T_1+T_2}{2}\).
Answer: (C) \(T=\dfrac{T_1+T_2}{2}\)
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