Physics/Option Modules/Engineering Physics/Refrigerators and Heat Pumps
Physics · Engineering Physics · A2 Level

Refrigerators and Heat Pumps

Heat always flows naturally from hot to cold — that's the whole basis of a heat engine. Refrigerators and heat pumps do the reverse: they spend work to force heat backward, from cold to hot, and closing out this topic means seeing how they fit into the same energy-flow accounting used throughout.

9 min read

Running the heat engine backward

A heat engine takes in heat from a hot source and delivers useful work, rejecting some waste heat to a cold sink along the way. A reversed heat engine — the basic principle behind both refrigerators and heat pumps — does the opposite: work $W$ is put into the system, and this is used to extract heat $Q_C$ from a cold reservoir and deliver a larger quantity of heat to a hot reservoir. Energy conservation fixes exactly how much heat is delivered to the hot side:

$$Q_H = Q_C + W$$

This is nothing more than the first law again, rearranged: the heat delivered to the hot reservoir is the heat drawn from the cold reservoir plus the work done to move it there. No knowledge of any specific practical refrigeration cycle or its components is needed here — what matters for this topic is purely this energy-flow accounting.

A schematic showing a cold reservoir bar labelled T-cold at the bottom and a hot reservoir bar labelled T-hot at the top, with a circle labelled reversed heat engine positioned between them. An arrow labelled Q-C points up from the cold reservoir into the device, an arrow labelled W-in points into the device from the side, and an arrow labelled Q-H points up from the device into the hot reservoir.
A reversed heat engine uses work input W to pump heat Q_C from a cold reservoir up to a hot reservoir, delivering Q_H = Q_C + W — the same energy-flow diagram as a heat engine, but with every arrow reversed.

Same physics, opposite purpose

A refrigerator and a heat pump are, at the level of this energy-flow picture, exactly the same device — the distinction between them is entirely about which of the two heat flows is considered the useful one.

  • Refrigerator. The useful effect is removing heat $Q_C$ from the cold space — keeping the inside of the fridge cold is the entire point. The heat $Q_H$ delivered to the hot reservoir (the room the fridge stands in, via the coils usually found at the back or underneath) is simply the unavoidable by-product of achieving that cooling.
  • Heat pump. The useful effect is delivering heat $Q_H$ to the hot space — warming a building is the entire point. The corresponding cooling of the cold reservoir it draws heat from (commonly the air outside, or the ground) is the by-product, of no direct interest in itself.

The same physical process, viewed from opposite ends, serves two entirely different practical purposes — which is also why the way "performance" is measured for each has to be defined around the flow that actually matters for that application.

Coefficient of performance

Because a refrigerator and a heat pump each care about a different one of the two heat flows, each has its own coefficient of performance (COP) — a ratio of the useful heat flow to the work input required to achieve it.

Worked ExampleCoefficients of performance for given reservoir temperatures
1
For a refrigerator, $COP_{\text{ref}} = \dfrac{Q_C}{W} = \dfrac{Q_C}{Q_H - Q_C} = \dfrac{T_C}{T_H - T_C}$. For a heat pump, $COP_{\text{hp}} = \dfrac{Q_H}{W} = \dfrac{Q_H}{Q_H - Q_C} = \dfrac{T_H}{T_H - T_C}$, with both temperatures in kelvin.
Each COP simply divides the useful heat flow for that device by the work input needed to produce it.
2
A heat pump draws heat from the ground at $T_C = 282\ \text{K}$ and delivers heat to a building's heating system at $T_H = 320\ \text{K}$. Find the maximum theoretical $COP_{\text{hp}}$, and the corresponding $COP_{\text{ref}}$ for the same two temperatures.
3
$COP_{\text{hp}} = \dfrac{T_H}{T_H - T_C} = \dfrac{320}{320 - 282} = \dfrac{320}{38} = 8.42$.
4
$COP_{\text{ref}} = \dfrac{T_C}{T_H - T_C} = \dfrac{282}{38} = 7.42$.
Notice $COP_{\text{hp}} = COP_{\text{ref}} + 1$ always, which follows directly from $Q_H = Q_C + W$ dividing through by $W$.
5
For every 1 J of work supplied to this heat pump, up to 8.42 J of heat is theoretically delivered to the building.
Result
Why COP greater than 1 doesn't break energy conservation

A coefficient of performance above 1 — often well above 1, as in the worked example — can look at first glance like it violates energy conservation: how can you get more energy out than you put in? The resolution is that COP is not a ratio of heat output to heat input; it's a ratio of useful heat delivered or removed to work input. The "extra" heat a heat pump delivers beyond the work supplied to it isn't created from nothing — it's heat that was already present in the cold reservoir (the ground, or the outside air) and has simply been moved, at the cost of the work input, to the hot reservoir instead. Energy is conserved throughout; what's remarkable is only how little work is needed to move a much larger quantity of heat that was already there.

Bringing the topic together

This closes the chain of ideas running through the whole of this topic. Internal energy set up the essential idea that, for an ideal gas, a temperature change is the signature of an energy change. The first law turned that into a precise, sign-sensitive statement of energy conservation for a gas being heated and doing work. Heat engines showed how that first law plays out physically inside a real engine cycle, using the p-V diagram and indicator diagrams to picture and measure the work done. The efficiency chain then tracked exactly how much of a fuel's chemical energy survives, stage by stage, all the way to a usable output shaft. The second law and the Carnot cycle revealed the hard ceiling on efficiency that no engine, however well designed, can ever exceed — and finally, reversed heat engines took the entire idea and ran it backward, using work to move heat against its natural direction, with refrigerators and heat pumps as two purposes built from exactly the same underlying process.

Key takeaways

  • A reversed heat engine uses work input $W$ to move heat from a cold reservoir to a hot reservoir, with $Q_H = Q_C + W$.
  • A refrigerator's useful effect is removing heat from the cold space; a heat pump's useful effect is delivering heat to the hot space — the same process, judged by opposite purposes.
  • $COP_{\text{ref}} = \dfrac{Q_C}{W} = \dfrac{Q_C}{Q_H-Q_C} = \dfrac{T_C}{T_H-T_C}$; $COP_{\text{hp}} = \dfrac{Q_H}{W} = \dfrac{Q_H}{Q_H-Q_C} = \dfrac{T_H}{T_H-T_C}$, with temperatures in kelvin.
  • Coefficients of performance can exceed 1, often substantially, since they relate useful heat moved to work input, not heat to heat — no violation of energy conservation is involved.
  • The full topic chain: internal energy → the first law → heat engines and the p-V/indicator-diagram picture → the efficiency chain → the Carnot ceiling from the second law → reversed engines running the same ideas backward.
  • No knowledge of specific practical refrigeration cycles or components is required — only this energy-flow accounting between the work input and the two heat flows.