Question

A carnot engine having an efficiency of 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)1J b)90J c)99J d)100J

Answer 1

the lower temperature is 1j

Answer 2

The energy absorbed from the reservoir at low temperature is 90J

Solution:

The given system has heat engine with an efficiency which is used as refrigerator, so the amount of energy absorbed can be obtained as follows,

Given: Efficiency = $\frac{1}{10}$

Efficiency, $\eta=1-\frac{Q_{2}}{Q_{1}}, 1-\eta=\frac{Q_{2}}{Q_{1}}$

Coefficient of refrigerator performance $\alpha=\frac{Q_{2}}{W}=\frac{Q_{2}}{Q_{2}-Q_{2}}=\frac{\frac{Q_{2}}{Q_{1}}}{1-\frac{Q_{2}}{Q_{1}}}=\frac{1-\eta}{\eta}$

$\begin{aligned} \alpha &=\frac{\left(1-\frac{1}{10}\right)}{\left(\frac{1}{10}\right)}=\frac{\left(\frac{9}{10}\right)}{\frac{1}{10}}=\frac{9}{10} \times 10=9 \\ \alpha &=\frac{Q}{W} \\ 9 &=\frac{Q}{10} \\ Q &=9 \times 10 \\ Q &=90 \mathrm{J} \end{aligned}$