# Refrigerator - Heat Energy

1. Nov 27, 2005

### dekoi

100 mL of water at 288 K is placed in the freezer compartment of a refrigerator with a coefficient of performance of 4.0. How much heat energy is exhausted into the room as the water is changed to ice at 258 K?

(The answer is 5.34 x 10^4 J)

I tried using the Brayton-cycle refrigerator to understand this problem.

I am assuming that the lower isobar is the part of the graph that I need to analyze, where the low-temperature heat exchanger warms the gas and cools the refrigerator. With this method, I know that $$Q = nC_v\Delta T$$. However, I cannot think of a way to find the number of moles (n).

I also don't know if my reasoning is correct (that is, only using the lower isobar of the graph to answer my question).

I am also assuming that my lower temperature (258 K) is equivalent to Tc, and my higher temperature (288 K) is equivalent to Th.
I am not sure if this is correct, however.

I have realized that I need to use work (in) to answer the question, because K = Q / W.

However, the work for an isobar is W = pV, and I do not know the pressure, and am unable to find it since I do not know the number of moles.

Any advice will be greatly appreciated, thank you.

Last edited by a moderator: Nov 27, 2005
2. Nov 27, 2005

### Andrew Mason

You don't know, or have to know, the details of the cycle. All you have to do is find $Q_H = W + Q_C$ where:
$$COP = Q_C/W$$

$$Q_C$$ is the heat lost by the water (you have to take into account temperature change + heat of fusion).

AM

Last edited: Nov 27, 2005
3. Nov 28, 2005

### dekoi

What is COP?

And I don't think we are required to know heat of fusion, so there must be some other way to solve the problem.

4. Nov 28, 2005

### Andrew Mason

You have double posted this. My response will be found on the other thread: