Two Carnot cycles have the same efficiency. The first operates between T_h = 200°C and T_c = 100°C and the second operates with a hot reservoir of T_h = 400°C. The lower reservoir's temperature is

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Quizplus · Accepted Answer

The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_c}{T_h}$, where $T_c$ and $T_h$ are the temperatures of the cold and hot reservoirs, respectively, in Kelvin. For the first engine, converting the given temperatures to Kelvin gives $T_{h1} = 473K$ and $T_{c1} = 373K$. The efficiency is $\eta_1 = 1 - \frac{373}{473}$.For the second engine with $T_{h2} = 673K$, and since both engines have the same efficiency, we set $\eta_2 = \eta_1$, leading to $1 - \frac{T_{c2}}{673} = 1 - \frac{373}{473}$. Solving for $T_{c2}$ gives $T_{c2} = 531K$, which corresponds to $258°C$ when converted back to Celsius.

Two Carnot Cycles Have the Same Efficiency