# Carnot Air Conditioner

1. Nov 28, 2004

### 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.

2. Nov 28, 2004

### Andrew Mason

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

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

AM

3. Nov 29, 2004

### 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.

4. Nov 29, 2004

### Andrew Mason

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

Last edited: Nov 30, 2004