Question

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 $10J$, the amount of energy absorbed from the reservoir at lower temperature is

• A. $90J$
• B. $99J$
• C. $100J$
• D. $1J$

Answer 1

The correct option is A $90J$

Given, $\eta =\frac{1}{10}$
$\eta =\frac{\text{Desired effect}}{\text{work input}}=\frac{W_{out}}{Q_2}=\frac{Q_2-Q_1}{Q_2}$
Where, $Q_1$ is the heat at higher temperature and $Q_2$ heat at lower temperature

In case of refrigerator $Q_2$ is desired and work input is $Q_2-Q_1=W$. Where, $W$ is the work done on the system
$\beta =\frac{\text{Desired effect}}{\text{work input}}=\frac{Q_1}{Q_2-Q_1}$
$\begin{array}{}\frac{1}{\beta }=\frac{Q_2}{Q_1}-1& \text{(1)}\end{array}$
$\eta =1-\frac{Q_1}{Q_2}=\frac{1}{10}$
$\frac{Q_1}{Q_2}=\frac{9}{10}$ put in equation 1 we get
$\frac{1}{\beta }=\frac{10}{9}-1=\frac{1}{9}\Rightarrow \beta =9$
$9=\frac{Q_1}{W}=\frac{Q_1}{10}$
$Q_1=90J$