A Carnot engine,having an efficiency of eta = | vedclass

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(C) The efficiency $(\eta)$ of a Carnot engine and the coefficient of performance $(\beta)$ of a refrigerator are related as:$\beta = \frac{1 - \eta}{

Vedclass · Accepted Answer

$90$

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(C) The efficiency $(\eta)$ of a Carnot engine and the coefficient of performance $(\beta)$ of a refrigerator are related as:$\beta = \frac{1 - \eta}{\eta}$Given $\eta = 1/10$,we calculate the coefficient of performance:$\beta = \frac{1 - 1/10}{1/10} = \frac{9/10}{1/10} = 9$The coefficient of performance $(\beta)$ is also defined as the ratio of heat absorbed from the cold reservoir $(Q_2)$ to the work done $(W)$ on the system:$\beta = \frac{Q_2}{W}$Given $W = 10 \ J$ and $\beta = 9$,we have:$9 = \frac{Q_2}{10 \ J}$$Q_2 = 9 	imes 10 \ J = 90 \ J$Therefore,the energy absorbed from the reservoir at a lower temperature is $90 \ J$.

$A$ Carnot engine,having an efficiency of $\eta = 1/10$ as a heat engine,is used as a refrigerator. If the work done on the system is $10 \ J$,the amount of energy absorbed from the reservoir at a lower temperature is ....... $J$.

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