# Homework Help: Thermodynamics in a refridgerator

1. Feb 19, 2013

### Saxby

1. The problem statement, all variables and given/known data
A refrigerator operating in a Carnot cycle is used to create 273 K by abstracting heat from water at 273 K and giving out heat to a room at 300K. To produce 3 kg of ice it abstracts 1.0 * 106J of heat from the water.

Calculate:
(i) The amount of heat given out to the room;
(ii) The amount of work supplied by the refrigerator;
(iii) The coefficient of performance of the refrigerator.

2. Relevant equations
Q = m*c*ΔT
W = Q = Qc - Qh
e = Qc / (Qh - Qc)

Qc = Heat in cold reservoir
Qh = Heat in hot reservoir

3. The attempt at a solution
I find the phrasing of this questions quite strange, what does it mean by 'used to create 273K' and is 300K just the temperature of the room? I suppose the heat given out would be 1.0 * 106J + W but i do i find W??? (Just to point out we're not given the specific heat capacity of water or air)

2. Feb 19, 2013

### Staff: Mentor

The cold side is at 273K, and the hot side (room) is at 300K.
I think you have to assume a perfect Carnot engine here - this allows to calculate the (minimal) W required. You do not need any specific heat capacity.

3. Feb 19, 2013

### Andrew Mason

You do not need to know the specific heat of water, as mfb pointed out. You just need to know how much heat is removed and the fact that it is a Carnot cycle.

You have provided the expression for the COP (which you refer to as e) but you have to relate this to temperatures. (hint: what is the COP for a Carnot refrigerator operating between these temperatures?)

Your expression for W is not quite right. W = |Qh|-|Qc|, where W is the work done ON the system. You can express W in terms of Qc and COP.

W is determined from the COP and Qc. You can determine Qh from that as well.

AM