A Carnot air conditioner has a coefficient of performance of 17.0 and removes 72.0 MJ of thermal energy from the interior of a house every hour.How much power does it consume?

Question

Quizplus · Accepted Answer

The coefficient of performance (COP) for a Carnot air conditioner is given by the formula COP = Qc/W, where Qc is the amount of heat removed from the cold reservoir (in this case, the interior of the house) and W is the work input. Given COP = 17.0 and Qc = 72.0 MJ per hour, we can rearrange the formula to solve for W: W = Qc / COP. First, convert Qc to the same time unit as the power we're looking for (Watts, which is Joules per second). 72.0 MJ = 72,000,000 Joules, and since there are 3600 seconds in an hour, the power consumed is W = (72,000,000 J / 3600 s) / 17 = 1,176,470.59 W, which rounds to approximately 1180 W.

A Carnot Air Conditioner Has a Coefficient of Performance of 17.0