Two Carnot engines A and B operate in series | vedclass

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(A) For Carnot engine $A$: Efficiency $\eta_A = 1 - \frac{T}{T_1} = \frac{W_A}{Q_1}$. Thus,$W_A = Q_1 \left(1 - \frac{T}{T_1}\right) = Q_1 - \frac{Q_1

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$\frac{2}{3} T_1 + \frac{1}{3} T_3$

Vedclass · Answer

(A) For Carnot engine $A$: Efficiency $\eta_A = 1 - \frac{T}{T_1} = \frac{W_A}{Q_1}$. Thus,$W_A = Q_1 \left(1 - \frac{T}{T_1}\right) = Q_1 - \frac{Q_1 T}{T_1}$.Since $\frac{Q}{Q_1} = \frac{T}{T_1}$,we have $Q = \frac{Q_1 T}{T_1}$.For Carnot engine $B$: It absorbs heat $Q_B = Q/2$ at temperature $T$ and rejects heat $Q_3$ at temperature $T_3$. Efficiency $\eta_B = 1 - \frac{T_3}{T} = \frac{W_B}{Q_B}$.Thus,$W_B = \frac{Q}{2} \left(1 - \frac{T_3}{T}\right) = \frac{Q}{2} - \frac{Q T_3}{2 T}$.Given $W_A = W_B$,so $Q_1 - Q = \frac{Q}{2} - Q_3$.Substituting $Q = \frac{Q_1 T}{T_1}$,we get $Q_1 - \frac{Q_1 T}{T_1} = \frac{Q_1 T}{2 T_1} - Q_3$.Rearranging terms: $Q_1 + Q_3 = \frac{Q_1 T}{T_1} + \frac{Q_1 T}{2 T_1} = \frac{3 Q_1 T}{2 T_1}$.Using the relation for engine $B$,$\frac{Q_3}{Q/2} = \frac{T_3}{T} \Rightarrow Q_3 = \frac{Q T_3}{2 T} = \frac{Q_1 T T_3}{2 T_1 T} = \frac{Q_1 T_3}{2 T_1}$.Substituting $Q_3$ back: $Q_1 + \frac{Q_1 T_3}{2 T_1} = \frac{3 Q_1 T}{2 T_1}$.Dividing by $Q_1$: $1 + \frac{T_3}{2 T_1} = \frac{3 T}{2 T_1}$.Multiplying by $\frac{2 T_1}{3}$: $T = \frac{2 T_1}{3} + \frac{T_3}{3} = \frac{2}{3} T_1 + \frac{1}{3} T_3$.

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