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 thermal energy from the interior of the house?

Question

Quizplus · Accepted Answer

The efficiency of a Carnot refrigerator or air conditioner is given by the formula: COP = Qc/W = Tc / (Th - Tc), where Qc is the heat removed from the cold reservoir (interior of the house), W is the work input, Tc is the temperature of the cold reservoir in Kelvin, and Th is the temperature of the hot reservoir in Kelvin. First, convert the temperatures from Celsius to Kelvin: Tc = 20°C + 273.15 = 293.15 K, Th = 39°C + 273.15 = 312.15 K. Then, rearrange the formula to solve for W: W = Qc / COP = Qc / (Tc / (Th - Tc)). Plugging in the values: W = 2000 J / (293.15 / (312.15 - 293.15)) = 2000 J / (293.15 / 19) ≈ 130 J.

An Ideal Carnot Air Conditioner Operates Between an Indoor Temperature