A Carnot engine having an efficiency of $\frac{1}{10}$ is being used as a refrigerator. If the work done on the refrigerator is $10 \mathrm{~J}$, the amount of heat absorbed from the reservoir at lower temperature is:
Correct Option: , 4
(4) For carnot refrigerator
Efficiency $=\frac{Q_{1}-Q_{2}}{Q_{1}}$ Where,
$Q_{1}=$ heat lost from sorrounding
$Q_{2}=$ heat absorbed from reservoir at low temperature.
Also, $\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{w}{Q_{1}}$
$\Rightarrow \frac{1}{10}=\frac{\mathrm{w}}{\mathrm{Q}_{1}}$
$\Rightarrow Q_{1}=w \times 10=100 J$
So, $Q_{1}-Q_{2}=w$
$\Rightarrow Q_{2}=Q_{1}-w$
$\Rightarrow 100-10=Q_{2}=90 J$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.