# Thermodynamics: Refrigerator

1. May 24, 2010

### irycio

1. The problem statement, all variables and given/known data
In an isolated room there is a refrigerator. The refrigerator consumes electric power P=50W.
Mass of the air in a room m=500 kg, and it's specific heat capacity c=714 J/kg. Calculate the stream of heat (dQ/dT) from the refrigerator assuming it's stationary way of work and how will the temperature in the room increase after t'=1h.

2. Relevant equations
$$P=\frac{W}{t}$$
$$Q=c m \Delta T$$
1st law of thermodynamics

3. The attempt at a solution

Now, the exercise seems to be extremely simple or I completely misunderstand it.
Due to the stationary way of work, we expect the temperature inside the refrigerator to remain constant. Thus, the heat entering it should be equal one leaving it, and all the power should be used for this heat transfer. Hence:
W=Q (since the total change of energy should be 0)
$$\frac{dW}{dt}=\frac{dQ}{dt}$$
$$\frac{dQ}{dt}=P=50W=\frac{Q}{t}$$

And regarding the increase in the temperature:

$$Q=cm \Delta T$$
$$\Delta T=\frac{Q}{cm}=\frac{P*t'}{cm}=\frac{50*3600}{500*714} \approx 0.5 K$$

Which seems to be quite a reasonable result.

Anyway, please correct me if I'm wrong. I mean, I was thinking of using Carnot's cycle for this exercise, but that doesn't seem to apply to this situation, as we assume the stationary mode of work. Furthermore we are not wprovided with ANY data necessary, like the temperature or thermoinsulation of the refrigerator or whatever.

2. May 24, 2010

### Andrew Mason

Re: Thermodynamics-Refrigerator

Your analysis is correct. The rate at which heat is removed from the fridge is the same as the rate at which heat enters the fridge (the question omits the complexities of an actual fridge which uses a thermostat).

So the heat delivered to the room is the Qh of the refrigerator less the Qc that leaks back and is subsequently removed.

From the first law: Qh = Qc + W

Net heat contributed to the room, therefore, is Qh-Qc = W = 50 W.

AM