# 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

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

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

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