HVAC & Refrigeration
Intermediate

Coefficient of Performance (Cooling)

Cooling delivered per unit of work input — higher is more efficient.

Formula

COP=QcWCOP = \frac{Q_c}{W}

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Variables

COPCoefficient of performance
Q_cHeat removed (W)
WWork input (W)

Example

Q_c=3500 W, W=1000 W: COP = 3.5

How it works

The coefficient of performance for cooling tells you how much heat a refrigeration or air-conditioning system removes from a cold space for each unit of work it consumes. It works because a refrigerator does not create cold; it uses input work to move thermal energy from inside the cooled space to the warmer surroundings. Since the useful output in cooling is the heat removed from the space, a larger COP means the system delivers more cooling for the same electrical power.

When to use it

Use this when evaluating refrigerators, freezers, air conditioners, chillers, or heat pumps operating in cooling mode. It is especially useful for comparing energy efficiency across different systems or estimating how much cooling you get from a known power input.

Worked example, step by step

Given: Q_c = 4800 W, W = 1500 W Step 1: Write the formula: COP = Q_c / W Step 2: Substitute the values: COP = 4800 / 1500 Step 3: Calculate: COP = 3.2 Final answer: The coefficient of performance is 3.2, meaning the system removes 3.2 watts of heat for every 1 watt of work input.

Common mistakes to avoid

  • Using the heat rejected to the room instead of the heat removed from the cooled space
  • Putting temperature values into the formula instead of heat flow and work
  • Forgetting that COP is a ratio and has no units

Did You Know?

A COP of 3.5 means the unit moves 3.5 times more heat than the electrical energy it consumes.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026