An ideal heat engine has an efficiency eta. T | vedclass

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(C) For an ideal heat engine,efficiency $\eta$ is given by:$\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}$F

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$\left(\frac{1}{\eta}\right)-1$

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(C) For an ideal heat engine,efficiency $\eta$ is given by:$\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}$From this,we get:$\frac{T_2}{T_1} = 1 - \eta$$\frac{T_1}{T_2} = \frac{1}{1 - \eta} \quad \dots (i)$When the heat engine is operated in the backward direction,it acts as a refrigerator. The coefficient of performance $\beta$ is given by:$\beta = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2}$Dividing the numerator and denominator by $T_2$:$\beta = \frac{1}{\frac{T_1}{T_2} - 1}$Substituting the value from equation $(i)$:$\beta = \frac{1}{\frac{1}{1 - \eta} - 1} = \frac{1}{\frac{1 - (1 - \eta)}{1 - \eta}} = \frac{1 - \eta}{\eta} = \frac{1}{\eta} - 1$

An ideal heat engine has an efficiency $\eta$. The coefficient of performance of the engine when driven backward will be

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