GATE 2021 PI – Question 32

General Engineering · Thermal and Fluids Engineering · 1 mark · Numerical answer

A refrigerator removes heat at 300 kJ/min to maintain its inside at 2°C. Input power is 2 kW. Find COP to one decimal.

Show answer and explanation

Correct answer: 2.4 to 2.6

Explanation

Cooling rate = 300/60 = 5 kW. Refrigerator COP = heat removed/input work = 5/2 = **2.5**.

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