Q.An ideal refrigerator keeps its contents at 0°C while the room temperature is 27°C. Calculate its coefficient of performance.

A refrigerator does not "create cold." It moves heat from a cold place (the inside of the fridge) to a hot place (the kitchen). That is the first thing to hold in your mind. The cold side gets colder because heat is being pulled out of it; the hot side gets hotter because that same heat is dumped there, plus the energy used to do the work.

If you put a hot bowl of soup inside the fridge, the fridge has to work harder. Why? Because more heat needs to be moved out. The "performance" of a fridge is simply: how much heat did I manage to pull out of the cold space, for every unit of work (electricity) I put in?

That is the intuition. You want a big "heat removed" for a small "work paid."

The precise statement

Let:

• $Q_c$ = heat extracted from the cold reservoir (the inside of the fridge). This is the "useful" effect.
• $Q_h$ = heat dumped into the hot reservoir (the kitchen). This is always larger than $Q_c$.
• $W$ = work input (electrical energy consumed by the compressor).

From the first law of thermodynamics (energy conservation), for a complete cycle:

The work you put in ends up as extra heat added to the hot side. So the Coefficient of Performance (COP) of a refrigerator is defined as:

Using $W = Q_h - Q_c$, we get the standard form:

A concrete example

Suppose a fridge extracts 200 J of heat from the inside ($Q_c = 200$ J) and dumps 250 J into the kitchen ($Q_h = 250$ J). Then the work done is $W = 250 - 200 = 50$ J.

This means: for every 1 J of electrical energy you pay for, the fridge moves 4 J of heat out of your food. That is a COP of 4 — quite good.

If the fridge were perfect (impossible), it would move heat with zero work, and COP would be infinite. Real fridges have COP values between 2 and 6, depending on the temperature difference they have to work against.

Why the formula makes physical sense …