An ideal Carnot air conditioner operates between an indoor temperature of 20°C and an outdoor temperature of 39°C. How much energy does it use to remove 2000 J of heat from the interior of
The house?
A) 340 J
B) 780 J
C) 105 J
D) 130 J
E) 520 J

Question

Quizplus · Accepted Answer

The efficiency of a Carnot air conditioner is determined by the temperatures of the hot and cold reservoirs. The formula for the coefficient of performance (COP) of a Carnot refrigerator is COP = Tc / (Th - Tc), where Tc and Th are the absolute temperatures (in Kelvin) of the cold and hot reservoirs, respectively. Converting the given temperatures to Kelvin: Tc = 20°C + 273 = 293 K, Th = 39°C + 273 = 312 K. The COP is 293 / (312 - 293) = 293 / 19 = 15.42. The work done (energy used) is the heat removed divided by the COP: 2000 J / 15.42 ≈ 130 J.

An Ideal Carnot Air Conditioner Operates Between an Indoor Temperature