# A Carnot engine having an efficiency

Question:

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:

1. $99 \mathrm{~J}$

2. $100 \mathrm{~J}$

3. $1 \mathrm{~J}$

4. $90 \mathrm{~J}$

Correct Option: , 4

Solution:

(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, $Q_{1}-Q_{2}=w$

$\Rightarrow Q_{2}=Q_{1}-w$

$\Rightarrow 100-10=Q_{2}=90 J$