Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Thermodynamics

Question:

A refrigerator works between 3°C and 33°C. It is required to remove 650 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is : (Take 1 cal = 4.2 Joules)

Options:

2.967 W

296.7 W

29.67 W

2967 W

Correct Answer:

296.7 W

Explanation:

COP of a refrigerator is given by 

COP = $\frac{Q_c}{W} = \frac{T_c}{T_h -T_c}$

$\Rightarrow W = Q_c \frac{T_h - T_c}{T_c} = 650\times 4.2 \times \frac{33-3}{273+3} = 2730\times \frac{30}{276} J = 296.7W$