# Second Law of Thermodynamics - heat pumps and air conditioners

## Homework Statement

A heat pump operates as an air conditioner, drawing 20000 Btu per hr from a building. It is operating between 21 degrees Celsius and 41 degrees Celsius.

a) If the COP (coefficient of performance) is 35% that of a Carnot air conditioner, what is the value of the effective COP (coefficient of performance)?

b) What is the value of the power (Watts) required for the compressor motor?

## Homework Equations

QL = heat removed from the low temperature area (inside the building)
QH = heat removed from the high temperature area (outside the building)
W = work done to remove the heat
TL = temperature of system at lower temperature (inside the building)
TH = temperature of system at higher temperature (outside the building)

COP = QL/W

COP = QL/(QH - QL)

COPideal = TL / (TH - TL)

## The Attempt at a Solution

I first converted Btu (British thermal unit) to joules: 1Btu = 1055.056 J. Then I tried finding the ideal COP by using the above equation with the given temperatures and got 1.05, but I'm pretty sure the ideal COP is not calculated this way. I tried multiplying 1.05 by 0.35 but got the wrong answer for part a. Thanks for any help.