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Question

Physics Question on Thermodynamics

A refrigerator works between 4C4^{\circ}C and 30C30^{\circ}C. It is required to remove 600calories600\, calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is : (Take1cal=4.2Joules)(Take\, 1 \,cal\, =\, 4.2 \,Joules)

A

23.65 W

B

236.5 W

C

2365 W

D

2.365 W

Answer

236.5 W

Explanation

Solution

β=Q2W=T2T1T2\beta = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2} (Where Q2Q_2 is heat removed)
600×4.2W=277303277\Rightarrow \, \frac{600 \times 4.2}{W} = \frac{277}{303 - 277}
W=236.5\Rightarrow \, W = 236.5 joule
Power=Wt=236.5joule1sec=236.5watt.\Rightarrow \, { Power = \frac{W}{t} = \frac{236.5 \,joule}{1 sec} = 236.5 \, watt.}