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

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{10}$.From this,we get $\frac{T_2}{T_1} = 1 - \frac{1}{10} = \

Vedclass · Accepted Answer

$90$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{1}{10}$.From this,we get $\frac{T_2}{T_1} = 1 - \frac{1}{10} = \frac{9}{10}$,which implies $\frac{T_1}{T_2} = \frac{10}{9}$.The Coefficient of Performance $(COP)$ of a refrigerator is given by $COP = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2} = \frac{1}{\frac{T_1}{T_2} - 1}$.Substituting the values,$COP = \frac{1}{\frac{10}{9} - 1} = \frac{1}{1/9} = 9$.Since $COP = \frac{Q_2}{W}$,where $Q_2$ is the heat absorbed from the cold reservoir and $W$ is the work done,we have $Q_2 = COP 	imes W$.Given $W = 10 \ J$,we get $Q_2 = 9 	imes 10 \ J = 90 \ J$.

$A$ 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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