Thermodynamics & Heat
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Coefficient of Performance

Efficiency measure of a refrigerator or heat pump.

Formula

COP=QCWCOP = \frac{Q_C}{W}

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Variables

Q_CHeat removed
WWork input

How it works

The coefficient of performance compares how much heat a refrigerator or heat pump can move to how much work it must be supplied. For a refrigerator, the useful effect is the heat removed from the cold space, so COP = Q_C / W; a larger value means the device transfers more thermal energy per unit of electrical or mechanical input. It works because refrigeration cycles do not create cold directly; they use work to move heat against its natural direction, and COP measures how effectively that work is used.

When to use it

Use this when analyzing refrigerators, freezers, air conditioners, or heat pumps in thermodynamics problems. It is especially useful when you are given heat extracted and work input, or when comparing the performance of different cooling or heating cycles.

Worked example, step by step

A laboratory freezer removes 18.0 kJ of heat from its interior while consuming 3.60 kJ of electrical work. Use COP = Q_C / W Substitute the values: COP = 18.0 kJ / 3.60 kJ Compute: COP = 5.00 Final answer: The freezer has a coefficient of performance of 5.00, meaning it removes 5 joules of heat for every 1 joule of work input.

Common mistakes to avoid

  • Using the heat released to the room instead of the heat removed from the cold space for refrigerator COP
  • Mixing up COP with thermal efficiency, which is limited to being less than 1 for heat engines but can exceed 1 for refrigerators and heat pumps
  • Forgetting that Q_C and W must be in the same energy units before taking the ratio

Did You Know?

A good refrigerator moves several times more heat than the electrical energy it consumes, beating 100% "efficiency".

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026