Question

In an refrigerator, compressor motor is of $1\text{kW}$ power. Heat is transferred from $–3^{\circ }C$ to ${27}^{\circ }C$. What amount of heat is coming out of the refrigerator per second assuming that efficiency of refrigerator is $50\%$ of a perfect engine?

• A. $5\text{kJ}$
• B. $6\text{kJ}$
• C. $19\text{kJ}$
• D. $20\text{kJ}$

Answer 1

The correct option is D $20\text{kJ}$

Given sink temperature
$T_i=-3^{\circ }C=(273-3)K=270K$
Source temperature $T_f={27}^{\circ }C=(273+27)K=300K$
Efficiency of ideal engine
$\eta =1-\frac{T_i}{T_f}$
$\eta =1-\frac{270}{300}=\frac{1}{10}$
Efficiency of refrigerator ${\eta }^{\prime }=\frac{1}{2}\times \frac{1}{10}=\frac{1}{20}$
Using
${\eta }^{\prime }=\frac{W}{Q}\frac{1}{20}=\frac{1\text{kJ}}{Q}\Rightarrow Q=20\text{kJ}$