Question

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 1 kW power, and heat is transferred from $–3^{\circ }Cto{27}^{\circ }C,$ find the heat taken
out of the refrigerator per second assuming its efficiency is 50% of a perfect heat engine.

• A. 14 kJ
• B. 12 kJ
• C. 19 kJ
• D. 20 kJ

Answer 1

The correct option is C 19 kJ

Efficiency of a perfect engine working between $-3^{\circ }Cand{27}^{\circ }(i.e.,T_2=270Kand{T}_1=300K){\eta }_{engine}=1-\frac{T_2}{T_1}=1-\frac{270K}{300K}=0.1$
Since efficiency of the refrigerator $({\eta }_{ref}.)is50\%of{\eta }_{engine}\therefore {\eta }_{ref.}=0.5{\eta }_{engine}=0.05$
If $Q_1$ is the heat transferred per second at higher temperature by doing work W, then ${\eta }_{ref.}=\frac{W}{Q_1}or{Q}_1=\frac{W}{{\eta }_{ref.}}=\frac{1kj}{0.05}=20kJ$
Since ${\eta }_{ref}.Is0.05,$ heat removed from the refrigerator per second, i.e.,
$Q_2=Q_1-\eta ref.Q_1=Q_1(1-{\eta }_{ref}.)=20kJ(1-0.05)=19kJ$