Question

A carnot engine, having an efficiency of $\eta =\frac{1}{10}$ 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. 99 J
• B. 90 J
• C. 1 J
• D. 100 J

Answer 1

The correct option is B 90 J

$w=Q_2[\frac{T_1}{T_2}-1]$
$\eta =1-\frac{T_2}{T_1}=\frac{1}{10}=1-\frac{T_2}{T_1}\Rightarrow \frac{T_2}{T_2}=\frac{9}{10}$
$\therefore 10=Q_2(\frac{10}{9}-1)$
$\Rightarrow 10=Q_2\left(\frac{1}{9}\right)$
$\therefore Q_2=90$