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NEET PHYSICSThermodynamicsMedium

Question

The coefficient of performance of a refrigerator is 55. If the temperature inside freezer is 20C-20^\circ\text{C}, the temperature of the surroundings to which it rejects heat is

A

31C31^\circ\text{C}

B

41C41^\circ\text{C}

C

11C11^\circ\text{C}

D

21C21^\circ\text{C}

Step-by-Step Solution

The coefficient of performance (β\beta) of a refrigerator is given by the formula: β=T2T1T2\beta = \frac{T_2}{T_1 - T_2} where T2T_2 is the absolute temperature of the cold reservoir (freezer) and T1T_1 is the absolute temperature of the hot reservoir (surroundings). Given: β=5\beta = 5 T2=20C=27320=253 KT_2 = -20^\circ\text{C} = 273 - 20 = 253\text{ K} Substituting these values into the formula: 5=253T12535 = \frac{253}{T_1 - 253} 5(T1253)=2535(T_1 - 253) = 253 5T11265=2535T_1 - 1265 = 253 5T1=15185T_1 = 1518 T1=15185=303.6 KT_1 = \frac{1518}{5} = 303.6\text{ K} Converting T1T_1 back to Celsius: T1(C)=303.6273=30.6C31CT_1 (^\circ\text{C}) = 303.6 - 273 = 30.6^\circ\text{C} \approx 31^\circ\text{C} Therefore, the temperature of the surroundings is nearly 31C31^\circ\text{C}.

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.

PHYSICSThermodynamicscoefficientperformancerefrigeratortemperatureinside

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