Question

A refrigerator has a coefficient of performance of 3.0. (a) If it requires 200 J of work per cycle, how much heat per cycle does it remove from the cold reservoir?

a) 150 J
b) 66.7 J
c) 100 J
d) 75 J

Answer 1

Given a refrigerator with a coefficient of performance (COP) of 3.0 and a work input of 200 J per cycle, it will remove 600 J of heat per cycle from the cold reservoir, which does not match the provided options.

Step-by-step explanation:

The coefficient of performance (COP) for a refrigerator is a measure of its efficiency and is defined as the amount of heat removed from the cold reservoir divided by the work input. Given a COP of 3.0 and a work input of 200 J per cycle, we can calculate the amount of heat removed from the cold reservoir using the formula:

COP = Q_c/W

Where Q_c is the heat removed from the cold reservoir, and W is the work input. Rearranging the formula and solving for Q_c, we get:

Q_c = COP × W

Q_c = 3.0 × 200 J = 600 J

Therefore, the refrigerator removes 600 J of heat per cycle from the cold reservoir. Since the correct answer is not provided in the options, the details provided in the question might be insufficient or incorrect.