A Carnot engine having an efficiency of frac{ | vedclass

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(C) The efficiency of a Carnot engine is given by $\eta = \frac{W}{Q_H} = \frac{1}{10}$.When used as a refrigerator,the coefficient of performance $\b

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$90$

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(C) The efficiency of a Carnot engine is given by $\eta = \frac{W}{Q_H} = \frac{1}{10}$.When used as a refrigerator,the coefficient of performance $\beta$ is related to the efficiency $\eta$ of the engine by the relation $\beta = \frac{1 - \eta}{\eta}$.Substituting $\eta = \frac{1}{10}$,we get $\beta = \frac{1 - 1/10}{1/10} = \frac{9/10}{1/10} = 9$.By definition,the coefficient of performance is $\beta = \frac{Q_L}{W}$,where $Q_L$ is the heat absorbed from the lower temperature reservoir and $W$ is the work done.Given $W = 10 \; J$,we have $9 = \frac{Q_L}{10}$.Therefore,$Q_L = 9 	imes 10 = 90 \; J$.

$A$ Carnot engine having an efficiency of $\frac{1}{10}$ is being used as a refrigerator. If the work done on the refrigerator is $10 \; J,$ the amount of heat absorbed from the reservoir at lower temperature is .............. $J$.

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