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NEET PHYSICSThermodynamicsMedium

Question

A refrigerator works between 4C4^{\circ}\text{C} and 30C30^{\circ}\text{C}. It is required to remove 600 calories600 \text{ calories} of heat every second to keep the temperature of the refrigerated space constant. The power required will be: (Take, 1 cal=4.2 Joules1 \text{ cal} = 4.2 \text{ Joules})

A

23.65 W

B

236.5 W

C

2365 W

D

2.365 W

Step-by-Step Solution

The coefficient of performance (COP) or β\beta of a refrigerator working between temperatures T1T_1 (higher) and T2T_2 (lower) is given by: β=T2T1T2\beta = \frac{T_2}{T_1 - T_2} Given: T2=4C=277 KT_2 = 4^{\circ}\text{C} = 277 \text{ K} T1=30C=303 KT_1 = 30^{\circ}\text{C} = 303 \text{ K} β=277303277=2772610.65\beta = \frac{277}{303 - 277} = \frac{277}{26} \approx 10.65 Also, β\beta is defined as the ratio of heat extracted (Q2Q_2) to the work done (WW): β=Q2W\beta = \frac{Q_2}{W} Given rate of heat extraction Q˙2=600 cal/s\dot{Q}_2 = 600 \text{ cal/s}. Convert to Joules: Q˙2=600×4.2 J/s=2520 W\dot{Q}_2 = 600 \times 4.2 \text{ J/s} = 2520 \text{ W}. The power required (PP) corresponds to the work done per second (W˙\dot{W}): P=Q˙2β=252010.65P = \frac{\dot{Q}_2}{\beta} = \frac{2520}{10.65} P236.5 WP \approx 236.5 \text{ W}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermodynamicsrefrigeratorbetweencirctextccirctextcrequired

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