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 $10\text{J}$, the amount of heat absorbed from the reservoir at lower temperature is

• A. $99\text{J}$
• B. $100\text{J}$
• C. $1\text{J}$
• D. $90\text{J}$

Answer 1

The correct option is D $90\text{J}$

For carnot engine

$\text{Efficiency}=\frac{Q_1-Q_2}{Q_1}$

Where,

$Q_1=\text{heat lost from sorrounding}$

$Q_2=\text{heat absorbed from reservoir at low temperature.}$

$\text{Also,}\frac{Q_1-Q_2}{Q_1}=\frac{W}{Q_1}$

Given that, work done on the refrigerator is $W=10\text{J}$
$\Rightarrow \frac{1}{10}=\frac{W}{Q_1}$

$\Rightarrow Q_1=W\times 10=100\text{J}$

As $Q_1-Q_2=W$

$\Rightarrow Q_2=Q_1-W$


$\Rightarrow 100-10=Q_2=90\text{J}$