A Carnot refrigerator has a coefficient of performance of 2.5.The refrigerator consumes 50 W of power.How much heat is removed from the interior of the refrigerator in 1 hour?
A) 7.5 kJ
B) 450 kJ
C) 180 kJ
D) 720 kJ
E) 72 kJ

Question

Quizplus · Accepted Answer

The coefficient of performance (COP) for a refrigerator is defined as the ratio of the heat removed from the refrigerator (Q_c) to the work input (W). Given COP = 2.5 and W = 50 W, we can calculate Q_c as follows: COP = Q_c / W, so Q_c = COP * W = 2.5 * 50 W = 125 W. To find the heat removed in 1 hour, convert power (in watts) to energy (in joules) using the conversion 1 W = 1 J/s. Thus, 125 J/s * 3600 s = 450,000 J or 450 kJ.

A Carnot Refrigerator Has a Coefficient of Performance of 2