What Is the Coefficient of Performance of a Carnot Air Conditioner?

[BlackMamba]

Hi there,

I have what should be a simple problem, but it's driving me insane.

Here's the problem: A Carnot air conditioner maintains the temperature in a house at 290 K on a day when the temperature outside is 315 K. What is the coefficient of performance of the air conditioner?

So I know to find the coefficient it would be Qc/W. I have neither of those numbers in the problem. I do have Th and Tc and even if I do substitutions I will ALWAYS have two unknowns. At least the way I am attempting this.

My thinking was originally that I could substitute for Qc, since Oc = W + Qh but I still have two unknown variables... What am I forgetting??

Any help will be greatly appreciated.

 

[Andrew Mason]

Here's the problem: A Carnot air conditioner maintains the temperature in a house at 290 K on a day when the temperature outside is 315 K. What is the coefficient of performance of the air conditioner?

They expect you to determine the coefficient of performance of an ideal Carnot cycle using:

COP = \frac{T_C}{T_H-T_C}

AM

 

[BlackMamba]

Thanks for replying but how did you come to that point? In my book the equation for the coefficient of performance is Qc/W. No where in my book is your equation listed.

Thanks for your help.

 

[Andrew Mason]

Thanks for replying but how did you come to that point? In my book the equation for the coefficient of performance is Qc/W. No where in my book is your equation listed.

Thanks for your help.

For the refrigerator,
Q_H = \Delta W + Q_C
This means that the heat flowing out of the cold reservoir + work added is equal to the heat flowing into the hot reservoir.

For the reversible (ideal) Carnot cycle, \Delta S = 0 where S = Q/T
So:
Q_H/T_H - Q_C/T_C = 0

Combining these two equations:
Q_H = Q_C(T_H/T_C) = \Delta W + Q_C
T_H/T_C = \Delta W/Q_C + 1
Q_C/\Delta W = 1/(T_H/T_C - 1)

Which simply reduces to:
Q_C/\Delta W = T_C/(T_H-T_C)

AM