# Carnot Ideal Refrigerator

1. Dec 4, 2006

### swain1

1. The problem statement, all variables and given/known data
a) Assume that a Carnot cycle is used to achieve that 250kg of water at 0 degrees celsius is frozen into ice using an ideal refigerator The room temperature is 27 degrees. What is the minimum amount of work input to the refrigerator to achieve this?

b)Assume the refigerator works with 65% efficiency (compare to a carnot engine), how much work input does it require.

2. Relevant equations
Latent heat of melting ice =3.33E5 J/kg

3. The attempt at a solution
For part a) I don't see why the amount of work input is not just that required to turn the water to ice, which would be 3.33E5*250. I think I might be missing something as the room temperature is given. I haven't yet started the second part as I think I need to understand the first part. Any help would be appreciated.

2. Dec 4, 2006

### Andrew Mason

The amount of HEAT removed from the water has to be this amount. This is not the amount of work input, however.

The refrigerator has to remove heat from a 0 degree reservoir and deliver it as heat at 27 degrees. In order to do that, work must be input. If it was a Carnot cycle, the coefficient of performance would be:

$$COP = Q_c/W = Q_c/(Q_h-Q_c) = \frac{T_c/T_h}{1 - T_c/T_h}$$

In this case, what is the COP?

That will give you Qh and that will, in turn, give you W.

AM

3. Dec 4, 2006

### swain1

Ok I see that puting in the temps into the formula give the COP. Which I have found to be 10.1. I am not sure where this comes from though.
How am I meant to find the value of Qc so I can work out the value of W.

So for the first part I don't need the latent heat of melting ice? Thanks

4. Dec 4, 2006

### Andrew Mason

COP is Qc/W (the ratio of "cooling" or heat removal to work input). For a Carnot cycle heat flow is at constant temperature and $\Delta S = Q_h/T_h - Q_c/T_c = 0$ so Qc/Qh = Tc/Th
Qc is the heat removed from the water which you have correctly stated as the mass x latent heat/mass

AM

5. Dec 4, 2006

### swain1

Thanks for that. I understand that now. I got the answer 8.3E6 J which I hope is right. For part b) would the amount of power used just be (8.3E6)/0.65=1.3E7J or am I missing something?

6. Dec 4, 2006

### Andrew Mason

For part a) I get 8.23E6 J:

$$W = Q_c/COP = 3.33e6*250/(273/27) = 8.23e6 J$$

You have the right approach to b). If the real refrigerator (b) was 65% as efficient as the Carnot, it would require 1/.65 of the work needed in the Carnot cycle (COP would be .65 * COP of the Carnot). So the work would be 1.27e7 J.

AM