A Carnot engine works first between 200^{circ | vedclass

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Question

(A) Case $1$: The engine operates between $T_H = 200^{\circ} C = 473 \ K$ and $T_L = 0^{\circ} C = 273 \ K$.The efficiency is $\eta_1 = 1 - \frac{T_L}

Vedclass · Accepted Answer

$0.577$

Vedclass · Answer

(A) Case $1$: The engine operates between $T_H = 200^{\circ} C = 473 \ K$ and $T_L = 0^{\circ} C = 273 \ K$.The efficiency is $\eta_1 = 1 - \frac{T_L}{T_H} = 1 - \frac{273}{473} = \frac{200}{473} \approx 0.4228$.Case $2$: The engine operates between $T_H = 0^{\circ} C = 273 \ K$ and $T_L = -200^{\circ} C = 73 \ K$.The efficiency is $\eta_2 = 1 - \frac{T_L}{T_H} = 1 - \frac{73}{273} = \frac{200}{273} \approx 0.7326$.The ratio of efficiencies is $\frac{\eta_1}{\eta_2} = \frac{200/473}{200/273} = \frac{273}{473} \approx 0.577$.

$A$ Carnot engine works first between $200^{\circ} C$ and $0^{\circ} C$ and then between $0^{\circ} C$ and $-200^{\circ} C$. The ratio of its efficiency in the two cases is

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