# Carnot Refrigerator

1. Nov 28, 2006

### Kalie

A Carnot refrigerator operating between -10.0 C and 40.0 C extracts heat from the cold reservoir at the rate 300 J/s. What are (a) the coefficient of performance of this refrigerator, (b) the rate at which work is done on the refrigerator and (c) the rate at which heat is exhausted to the hot side?

A. All right for i got the answer to be 5.26 because
T_c/(T_h-T_c)=coefficient of performance
This is correct

For Part B and C
I am unsure how to approach this
please someone help me with an equation....

2. Nov 29, 2006

### Andrew Mason

The co-efficient of performance of a refrigerator is the ratio of heat removed from the cold reservoir to the work done:

$$\text{COP} =\frac{Q_c}{W} = \frac{Q_c}{Q_h-Q_c} = \frac{Q_c/Q_h} {1 - Q_c/Q_h}$$

So W = Qc/COP. Since it is a Carnot cycle: Qc/Qh = Tc/Th

To find Qh use W = Qh-Qc

AM