**Question:**

**In a refrigerator, one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1kW power, and heat is transferred from -3oC to 27oC, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.**

**Solution:**

Efficiency of the Carnot engine is

$\eta=1-\frac{T_{2}}{T_{1}}$

Where

T1 = 300 K

T2 = 270 K

It is given that efficiency is 50% = 0.5

Efficiency of refrigerator = 0.05

We know that

$\eta_{\text {ref }}=\frac{W}{Q_{1}} \Rightarrow Q_{1}=\frac{W}{\eta_{\text {ref }}}=20 k J$

Therefore, the efficiency of the refrigerator is

$Q_{2}=Q_{1}-\eta_{\text {ref }} Q_{1}=Q_{1}\left(1-\eta_{\text {ref }}\right)=19 k J$