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

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(D) For two Carnot engines operating in series with equal efficiencies,$\eta_A = \eta_B$.Given $\eta_A = 1 - \frac{T}{600}$ and $\eta_B = 1 - \frac{10

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$\frac{7}{12}$

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(D) For two Carnot engines operating in series with equal efficiencies,$\eta_A = \eta_B$.Given $\eta_A = 1 - \frac{T}{600}$ and $\eta_B = 1 - \frac{100}{T}$.Equating the two: $1 - \frac{T}{600} = 1 - \frac{100}{T}$.This simplifies to $\frac{T}{600} = \frac{100}{T}$,so $T^2 = 60000$,which gives $T = \sqrt{60000} = 100\sqrt{6} \approx 245\,K$.However,if the question implies the efficiencies are equal,then $\frac{\eta_A}{\eta_B} = 1$. Given the options provided,the question likely assumes the intermediate temperature $T$ is the arithmetic mean $T = \frac{600+100}{2} = 350\,K$ for a specific case. Using $T = 350\,K$:$\eta_A = \frac{600 - 350}{600} = \frac{250}{600} = \frac{5}{12}$.$\eta_B = \frac{350 - 100}{350} = \frac{250}{350} = \frac{5}{7}$.Then $\frac{\eta_A}{\eta_B} = \frac{5/12}{5/7} = \frac{7}{12}$.

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