A Carnot engine, having an efficiency of $\eta = 1/10$ as 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 lower temperature is ....... $J$
AIEEE 2007,NEET 2017,AIPMT 2015,JEE MAIN 2020, Diffcult
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The efficiency $\left( \eta \right)$ of a Carnot engine and the coefficient of performance $\left( \beta \right)$ of a refrigerator are related as
$\beta = \frac{{1 - \eta }}{\eta }$ $Here,\,\eta = \frac{1}{{10}}$ $\therefore \beta \frac{{1 - \frac{1}{{10}}}}{{\left( {\frac{1}{{10}}} \right)}} = 9.$
Also, Coefficient of performance $\left( \beta \right)$ is given by $\beta = \frac{{{Q_2}}}{W}$
where $Q_2$ is the energy absorbed from the reservoir.
or, $9 = \frac{{{Q_2}}}{{10}}$ $\therefore {Q_2} = 90\,J$
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