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# JEE Class main Answered

A refrigerator works between 4°c and 30°c. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerator space constant. The power required is [Take 1 cal=4.2 joules]
Asked by yashu22022006 | 25 May, 2024, 09:13: AM

Coefficient of performance COP of refregirator is

where TC = 4o C  = (273+4) K = 277 K is cold source temperature and

TH = 30o C = ( 273+30) K =  303 K is hot source temperature .

Hence COP of refregirator is

COP = 277 / ( 303 - 277 ) = 277 / 26 = 10.65

If QC is heat energy removed from cold source and W is work done , then coefficient performance is

COP = QC / W

Hence Work done  W = QC / COP = [ ( 600 × 4.2 ) / 10.65 ] J = 237 J

Hence required power = 237 W

Answered by Thiyagarajan K | 25 May, 2024, 01:00: PM
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