A Carnot engine having an efficiency of 1/5 a | vedclass

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(C) The efficiency of a Carnot engine is given by $\eta = 1/5$. When this engine is used as a refrigerator,its coefficient of performance $\beta$ is r

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$200$

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(C) The efficiency of a Carnot engine is given by $\eta = 1/5$. When this engine is used as a refrigerator,its coefficient of performance $\beta$ is related to the efficiency $\eta$ by the formula: $\beta = \frac{1 - \eta}{\eta}$. Substituting the value of $\eta$: $\beta = \frac{1 - 1/5}{1/5} = \frac{4/5}{1/5} = 4$. The coefficient of performance is also defined as the ratio of heat absorbed from the cold reservoir $(Q)$ to the work done on the system $(W)$: $\beta = \frac{Q}{W}$. Given $W = 50 \ J$,we have $4 = \frac{Q}{50}$. Therefore,$Q = 4 	imes 50 = 200 \ J$.

$A$ Carnot engine having an efficiency of $1/5$ as a heat engine is used as a refrigerator. If the work done on the system is $50 \ J$,then the amount of energy absorbed from the reservoir at the lower temperature is (in $J$)

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