Question

A carnot engine having an efficiency $\eta =\frac{1}{10}$ 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:

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

Answer 1

The correct option is C $90J$

Efficiency of a carnot engine:

$\eta =1-\frac{Q_2}{Q_1}=\frac{Q_1-Q_2}{Q_1}=\frac{W}{Q_1}=\frac{10}{Q_1}$

$\therefore \frac{1}{10}=\frac{10}{Q_1}$

$\therefore Q_1=100J$

$\therefore Q_2=Q_1-W=100-10=90J$