Two Carnot engines A and B are operated in se | vedclass

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(D) For a Carnot engine,the work output $W$ is given by $W = Q_1 - Q_2 = Q_1(1 - T_2/T_1).$ Since the engines are in series,the heat rejected by engin

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$500$

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(D) For a Carnot engine,the work output $W$ is given by $W = Q_1 - Q_2 = Q_1(1 - T_2/T_1).$ Since the engines are in series,the heat rejected by engine $A$ is the heat received by engine $B.$ Let $Q_1$ be the heat input to $A,$ $Q_2$ be the heat rejected by $A$ (and received by $B$),and $Q_3$ be the heat rejected by $B.$ The work done by engine $A$ is $W_1 = Q_1 - Q_2 = Q_1(1 - T_2/T_1).$ The efficiency of a Carnot engine is $\eta = W/Q_{in} = 1 - T_{cold}/T_{hot}.$ Given that the work outputs are equal,$W_1 = W_2.$ For engine $A,$ $W_1 = Q_1 - Q_2.$ For engine $B,$ $W_2 = Q_2 - Q_3.$ Since $W_1 = W_2,$ we have $Q_1 - Q_2 = Q_2 - Q_3.$ Using the property of Carnot engines,$Q_1/T_1 = Q_2/T_2 = Q_3/T_3 = k.$ Thus,$Q_1 = kT_1,$ $Q_2 = kT_2,$ and $Q_3 = kT_3.$ Substituting these into the work equation: $kT_1 - kT_2 = kT_2 - kT_3.$ $T_1 - T_2 = T_2 - T_3 \Rightarrow 2T_2 = T_1 + T_3.$ $T_2 = (T_1 + T_3) / 2 = (600 + 400) / 2 = 1000 / 2 = 500 \, K.$

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