In 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 $-3^{0}Cto{27}^{0}C,$ find the heat taken out of the refrigerator per second assuming its efficiency is $50\%$ of a perfect engine.

Efficiency of a perfect engine working between $-3^{0}C$ and ${27}^{\circ }\text{C}$ (i.e., ${\text{T}}_2=270K$ and ${\text{T}}_1=300K$ )

 ${\text{η}}_{engine }=1-\frac{{\text{T}}_2}{{\text{T}}_1}=1-\frac{270\text{K}}{300\text{K}}=0.1$ 

Since efficiency of the refrigerator $\left({\text{η}}_{ref. }\right)$ is 50% of ${\text{η}}_{engine }$ 

$\therefore {\text{η}}_{ref. }=0.5{\text{η}}_{engine }=0.05$

(Using (i)) If Q₁ is the heat transferred per second at higher temperature by doing work W, then ${\text{η}}_{\text{ref.}}=\frac{\text{W}}{{\text{Q}}_1} or {\text{Q}}_1=\frac{\text{W}}{{\text{η}}_{\text{rec.}}}=\frac{1\text{kJ}}{0.05}=20\text{kJ }( as \text{W}=1\text{kW}\times 1\text{s}=1\text{kJ})$

Since ${\text{η}}_{ref }$ is 0.05, heat removed from the refrigerator per second, i.e.,

$\begin{array}{rr}{\text{Q}}_2& ={\text{Q}}_1-{\text{η}}_{ref. }{\text{Q}}_1={\text{Q}}_1(1-{\text{η}}_{ref. }) =20\text{kJ}(1-0.05)=19\text{kJ}\end{array}$