# A Carnot engine having an efficiency of  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 : Option 1) 1 J Option 2) 90 J Option 3) 99 J Option 4) 100 J

As we learnt in

Coefficient of performance (beta) -

$\beta = \frac{Q_{2}}{Q_{1}-Q_{2}}= \frac{T_{2}}{T_{1}-T_{2}}$

- wherein

$T_{1}> T_{2}$

For a perfect refrigerator $\beta \rightarrow \infty$

Efficiency of a carnot cycle -

$\eta =\frac{W}{Q_{1}-Q_{2}}=1-\frac{T_{2}}{T_{1}}$

$T_{1}\, and\, T_{2}$  are in kelvin

- wherein

$T_{1}=$ Source temperature

$T_{2}=$ Sink Temperature

$\left ( T_{1} > T_{2}\right )$

For a carnot cycle

$\beta =\frac{1}{\eta }-1$

$\beta$ = Coefficient of performance

$\eta$ = Efficiency of carnot engine

$\eta=\frac{1}{10}\:\:\:\:\:\:\:\:\:\:\:\:=> \beta =9$

$=> \beta=\frac{Q_{2}}{W}$

$Q_{2}=\beta.W=9\times 10 J=90J$

Option 1)

1 J

This option is incorrect.

Option 2)

90 J

This option is correct.

Option 3)

99 J

This option is incorrect.

Option 4)

100 J

This option is incorrect.

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