Temperature of a cold reservoir of a Carnot e | vedclass

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(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$,where $T_C$ is the temperature of the cold reservoir and $T_H$ i

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$227$

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(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$,where $T_C$ is the temperature of the cold reservoir and $T_H$ is the temperature of the hot reservoir in Kelvin.Given: $T_C = 127^{\circ} C = 127 + 273 = 400 \ K$ and $\eta = 20 \% = 0.2$.Substituting the values into the formula: $0.2 = 1 - \frac{400}{T_H}$.Rearranging the terms: $\frac{400}{T_H} = 1 - 0.2 = 0.8$.Solving for $T_H$: $T_H = \frac{400}{0.8} = 500 \ K$.Converting back to Celsius: $T_H = 500 - 273 = 227^{\circ} C$.

Temperature of a cold reservoir of a Carnot engine is $127^{\circ} C$. If the efficiency of the Carnot engine is $20 \%$,then the temperature of the hot reservoir is (in $^{\circ} C$)

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