# Homework Help: COP refrigerator/freezer problem

1. Oct 6, 2011

### Pauli_Pocket

1. The problem statement, all variables and given/known data
Compare a refrigerator set at +2 oC and a freezer set at -18 oC, both of which are operating in the room with temperature of +20 oC. What is relative amount of electric energy that will need to be used in both cases to cool a cup of water (250 ml) from room temperature to +5 oC? Assume ideal operation of both systems.

2. Relevant equations

C.O.P = Tcold/(Thot-Tcold)
C.O.P = Qc/w
Q = mCΔT

3. The attempt at a solution

C.O.P (refrigerator) = 275K/(293K-275K) = 15.278
C.O.P (freezer) = 255K/(293K-255K) = 6.711

Q = 250g(4.1855 J/g*K)(278-293K) = -15.696kJ

2. Oct 7, 2011

### qtm912

I would be tempted to say you are almost there and so all you need to do is to apply you second equation (in each case ) using the cop for thereby calculating w since you have worked out the Q cold value.

3. Oct 7, 2011

### Pauli_Pocket

If I just applied the 3rd equation I would be calculating work in kJ and the amount of work for a refrigerator to cool the water would be lower than for a freezer, which confuses me. Also the problem asks for electric energy in J/s or Watts.

(refrigerator) w = Qc/COP = -15.696 kJ/15.278 = -1.0274 kJ
(freezer) w = Qc/COP = -15.696 kJ/6.711 = -2.3388 kJ

4. Oct 7, 2011

### qtm912

I think the purpose of giving you the info abt the cup of water us to allow you to calculate the energy to be withdraw. You correctly calculated thus as 15.7 kJ. The fridge is more efficient with better cop, so you need just 15.7/15.278 kJ while the freezer needs 15.7/6.711 kJ.

The relationships look ok -- for pumps and refris lower temp differential better, for heat engines the opposite. Finally the qs as you have stated it is asking for energy not power so KJ is correct and not watts.
That's my take anyway.