Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Thermodynamics

Question:

A refrigerator works between 0°C and 30°C. It is required to remove 700 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:

2.365 W

290 W

236.5 W

323 W

Correct Answer:

323 W

Explanation:

COP of a refrigerator is given by

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

$\Rightarrow W = Q_c \frac{T_h - T_c}{T_c} = 700\times 4.2 \frac{30}{273} = 323W$