# Air conditioner homework

A home is kept cool by an air conditioner. The outside temperature is 311.75K and the interior of the home is 288.55K. If 127kJ/h of heat is removed from the house, what is the minimum power that must be provided to the air-conditioner? answer in kJ/h
my work

COP = Th / (Th - Tc)
= 311.75 / (311.75 - 288.55)
= 13.44

W = 127kJ/h / 13.44
= 9.45 kJ/h this is incorrect according to the homework server

Anybody know where i went wrong on this one???
any help is appreciated!

Sergio

## Answers and Replies

marcusl
Gold Member
There's no way the electrical input can be less than the power removed! You need to multiply, not divide.

Ok we went about this a different way and got the right answer...for those of you who may have the same problem...

Th, Tc and Qc are given in the problem...isolate Qh using Qh/Th = Qc/Tc
to find work
W = Qh-Qc

thank you for your help marcusl i realised i was going about it the wrong way :)

OlderDan
Homework Helper
There's no way the electrical input can be less than the power removed! You need to multiply, not divide.

I think that is not true. If the temperature difference goes to zero, it should ideally take no work to move the heat from one resevoir to another. I think the only problem with the original calculation is that the OP was using the COP for heating instead of the COP for cooling.

http://en.wikipedia.org/wiki/Coefficient_of_performance

Ok we went about this a different way and got the right answer...for those of you who may have the same problem...

Th, Tc and Qc are given in the problem...isolate Qh using Qh/Th = Qc/Tc
to find work
W = Qh-Qc

thank you for your help marcusl i realised i was going about it the wrong way :)

Qh/Th = Qc/Tc
Qh = Qc(Th/Tc)
W = Qh - Qc = Qc(Th/Tc) - Qc = Qc[(Th/Tc) - 1] = Qc(Th- Tc)/Tc = Qc/COP_cooling

COP_cooling = Tc/(Th- Tc) = 288.55/(311.75 - 288.55) = 12.44

W = 127kJ/h/12.44 = 10.21kJ/h

ahhh ok i see what i did wrong!

Perfect thank you very much for your time!!!

Sergio :D