# Carnot refrigerator

1. Apr 8, 2005

### sportsrules

A heat pump is essentially a refrigerator engine that uses the inside of a building as a hot reservoir in the winter and the outside of the building as the hot reservoir in the summer. A) For a carnot refrigerator engine operating between 25 C and 40 C in the summer, what is the coefficient of performance? B) For each joule of heat transfer form the cooler reservoir per cycle, how many joules of work are done by the carnot refrigerator engine?

Ok, I figured out part A, which was simply plugging the numbers into the equation

K=1/[(Thot/Tcold)-1]....and I got that K=19.9

I can't really figure out part B. I know that there is another equation

K=Qc/W
and
K=1/[(Qh/Qc)-1]

but I cannot figure out how to rearrange get an answer for the work b/c I don't know what Qc is or how to go about finding it. Any help would be great! Thanks!

2. Apr 9, 2005

### Andrew Mason

For a heat pump, the coeff. of perf. is the ratio of heat output (heat transferred from the cold reservoir to the hot reservoir) to the work input.

$$COP = \frac{Q_H}{W}$$

So:

$$W = Q_H/COP$$

$$COP = Q_H/W = Q_H/Q_H-Q_C = \frac{1}{1-\frac{Q_C}{Q_H}}$$

For the Carnot process (reversible so $\Delta S = 0$), the heat is exchanged at constant temperature, so:

$$\Delta S = Q_h/T_h - Q_c/T_c = 0$$

$$Q_h/T_h = Q_c/T_c$$

$$Q_h/Q_c = T_h/T_c$$

So: $$COP = \frac{1}{1-\frac{T_C}{T_H}}$$

COP = 1/(1-298/313) = 20.86

So for each Joule of heat delivered, W = 1/20.86 = .048 Joules of work input are needed

AM

Last edited: Apr 9, 2005