A Carnot engine has an efficiency of 25%. It | vedclass

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(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$,where $T_1$ is the temperature of the so

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$-33$

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(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.Given $\eta = 25\% = 0.25$ and $T_1 - T_2 = 80\,K$.Substituting the values: $0.25 = \frac{80}{T_1}$.Solving for $T_1$: $T_1 = \frac{80}{0.25} = 320\,K$.Now,find the temperature of the low-temperature reservoir $T_2$: $T_2 = T_1 - 80 = 320 - 80 = 240\,K$.To convert the temperature from Kelvin to Celsius: $T(^{\circ}C) = T(K) - 273.15$.$T_2 = 240 - 273.15 = -33.15^{\circ}C$,which is approximately $-33^{\circ}C$.

$A$ Carnot engine has an efficiency of $25\%$. It operates between reservoirs of constant temperature with a temperature difference of $80\,K$. What is the temperature of the low-temperature reservoir in $^oC$?

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