# Carnot air conditioner

## Homework Statement

A Carnot air conditioner takes energy from the thermal energy of a room at 70 degrees F and transfers it to the outdoors, which is at 96 degrees F. For each joule of electric energy required to operate the air conditioner, how many joules are removed from the room?

## Homework Equations

$$\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$$

## The Attempt at a Solution

I substituted W + QC for QH. Then I plugged in

W = 1,
TC = (70-32) * 4/5+273,
TH = (96-32) * 4/5+273

and solved for QC.

I get QC = 20.3 J, but the answer in the back is rounded 21 J. When I read the problem, I thought the question was asking for QC. Should the 1 J or work be included in the amount of joules being removed from the room?

Hello,

A Carnot cycle operating as a reverse heat pump (i.e. air conditioner) has a Coefficient of Performance (COP) this is similar to an efficiency but since it's value can be greater than 1 so it has the alternative name Coefficient of Performance. The COP can be defined as the ratio of the heat extracted from a low temperature source to the work done to accomplish that extraction.

For this application the COP can be calculated as

COP = 1 / ((Th / TL) - 1)

where Th is the higher temperature and TL is the lower temperature.

Then you can use this relation

COP = QL / W