A heat engine running backward
A heat engine absorbs heat \(Q_1\) from a hot reservoir, converts part of it into work \(W\), and rejects the remainder \(Q_2\) to a cold reservoir. Reverse every arrow and you have a refrigerator or a heat pump. NCERT puts it plainly: a refrigerator is a heat engine working in the opposite direction. Work \(W\) is now an input. The machine uses that work to extract heat \(Q_2\) from a cold body and deposit heat \(Q_1\) into a hot body.
This direction does not happen on its own. Left alone, heat flows from hot to cold; never the reverse. To drive heat the wrong way — from the cold interior of a fridge out into a warmer kitchen — the device must be fed work, normally as electrical energy through a compressor. That requirement is not an engineering inconvenience; it is a statement of the second law, which we reach at the end of this page.
The energy-flow diagram
Over one complete cycle the working substance returns to its starting state, so its internal energy is unchanged: \(\Delta U = 0\). Energy conservation then fixes the bookkeeping. The device receives \(Q_2\) from the cold reservoir and \(W\) from the surroundings, and rejects \(Q_1\) to the hot reservoir. Energy in equals energy out,
$$Q_1 = Q_2 + W \qquad\Longleftrightarrow\qquad W = Q_1 - Q_2.$$
The heat dumped into the hot reservoir always exceeds the heat pulled from the cold one, by exactly the work supplied. A fridge therefore warms the kitchen by more than it cools its own interior; a room is never cooled by leaving the refrigerator door open.
Coefficient of performance of a refrigerator
A refrigerator is judged not by how much work it consumes but by how much heat it removes per unit of work. That ratio is the coefficient of performance, written \(\alpha\) (also \(K\) or COP):
$$\alpha_{\text{fridge}} = \frac{\text{heat extracted from cold reservoir}}{\text{work supplied}} = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}.$$
The "benefit" of a refrigerator is the cooling, so \(Q_2\) sits on top. A larger \(\alpha\) means more cooling for the same electricity bill. Unlike the efficiency of an engine, which is bounded above by 1, \(\alpha\) carries no such cap — a point we return to below.
A refrigerator removes 250 J of heat from its interior for every 50 J of electrical work supplied to the compressor. Find (a) the heat rejected to the room and (b) the coefficient of performance.
(a) Heat rejected. By energy conservation over a cycle, \(Q_1 = Q_2 + W = 250 + 50 = 300~\text{J}\). The room receives 300 J — more than the 250 J pulled from the cold interior.
(b) COP. \(\alpha = \dfrac{Q_2}{W} = \dfrac{250}{50} = 5\). For each joule of work the refrigerator shifts 5 J of heat out of the cold space. The number exceeds 1, which is normal for a refrigerator and is not a violation of any law.
Refrigerator versus heat pump
A heat pump is the very same machine; only the useful product changes. A refrigerator is installed to cool the cold reservoir, so the prized quantity is \(Q_2\). A heat pump is installed to warm the hot reservoir — a room in winter, with the cold reservoir being the chilly outside air — so the prized quantity is \(Q_1\). The definition of COP follows the goal.
| Aspect | Refrigerator | Heat pump |
|---|---|---|
| Purpose | Cool the cold reservoir (interior) | Warm the hot reservoir (room) |
| Useful effect | Heat extracted, \(Q_2\) | Heat delivered, \(Q_1\) |
| COP definition | \(\alpha = \dfrac{Q_2}{W} = \dfrac{Q_2}{Q_1-Q_2}\) | \(\alpha' = \dfrac{Q_1}{W} = \dfrac{Q_1}{Q_1-Q_2}\) |
| Ideal (Carnot) value | \(\dfrac{T_2}{T_1-T_2}\) | \(\dfrac{T_1}{T_1-T_2}\) |
| Energy balance | Q1 = Q2 + W for both | |
Because both COPs share the same denominator \(W = Q_1 - Q_2\), and the numerators differ by exactly \(W\), the two are linked by a clean identity:
$$\alpha'_{\text{heatpump}} = \frac{Q_1}{W} = \frac{Q_2 + W}{W} = \frac{Q_2}{W} + 1 = \alpha_{\text{fridge}} + 1.$$
The heat-pump COP always exceeds the refrigerator COP by precisely one. A heat pump delivering heat to a room is therefore never worse than an electric heater, which would convert work to heat with a COP of 1; the pump's COP is at least 1 even at its limit, and is typically several times that.
Ideal (Carnot) COP
The highest COP between two fixed temperatures belongs to the reversible Carnot cycle run backward. NCERT establishes that for a Carnot cycle the heats exchanged satisfy the universal relation
$$\frac{Q_1}{Q_2} = \frac{T_1}{T_2},$$
with temperatures in kelvin. Substituting into the refrigerator definition gives the ideal limit. Dividing numerator and denominator of \(\alpha = Q_2/(Q_1-Q_2)\) by \(Q_2\) and using \(Q_1/Q_2 = T_1/T_2\),
$$\alpha_{\text{Carnot, fridge}} = \frac{T_2}{T_1 - T_2}, \qquad \alpha_{\text{Carnot, heatpump}} = \frac{T_1}{T_1 - T_2}.$$
Three readings follow. As the two temperatures approach each other, \(T_1 - T_2 \to 0\) and the COP diverges — pumping heat across a small gap is nearly free. As the cold reservoir is driven far below the hot one, \(T_2 \ll T_1\), the COP collapses, which is why deep refrigeration is costly. And no real refrigerator between the same two temperatures can beat this Carnot value.
The reversed cycle that sets this ceiling is the forward Carnot engine — see Carnot engine & efficiency for the full cycle and the \(\eta = 1 - T_2/T_1\) derivation.
Why COP can exceed 1
The most persistent confusion is treating COP as if it were efficiency. Efficiency of an engine, \(\eta = W/Q_1\), is a fraction of an input that is converted to work, and it must be less than 1 because \(W < Q_1\). The coefficient of performance is a different ratio entirely: a benefit per cost, \(Q_2/W\) or \(Q_1/W\), where the benefit can be much larger than the cost. A household refrigerator routinely has \(\alpha\) between 2 and 6; a heat pump's COP can sit between 3 and 5.
Running a Carnot engine backward makes the link to efficiency exact. With \(\eta = W/Q_1\) and \(Q_1 = Q_2 + W\),
$$\alpha_{\text{fridge}} = \frac{Q_2}{W} = \frac{Q_1 - W}{W} = \frac{1}{\eta} - 1 = \frac{1 - \eta}{\eta}.$$
A Carnot engine of efficiency \(\eta = 1/10\), used as a refrigerator, has \(\alpha = (1 - 0.1)/0.1 = 9\). A small efficiency makes a large COP — the same hardware that converts little heat into work, when reversed, moves a great deal of heat per unit work.
The Clausius statement of the second law
The refrigerator is the device behind the Clausius statement of the second law. NCERT gives it verbatim:
No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.
Heat does move from cold to hot inside a refrigerator — but not as the sole result. External work \(W\) is supplied and heat \(Q_1\) is rejected to the surroundings. The statement therefore outlaws a perfect refrigerator: one that would transfer heat up the gradient with no work input, \(W = 0\). Such a machine would have \(\alpha = Q_2/W \to \infty\). The Clausius statement is thus equivalent to saying the coefficient of performance of a refrigerator can never be infinite. It is, NCERT notes, completely equivalent to the Kelvin–Planck statement that no engine can be perfectly efficient.
Refrigerators & heat pumps in one breath
- A refrigerator/heat pump is a heat engine in reverse: work \(W\) in, heat \(Q_2\) drawn from the cold reservoir, heat \(Q_1\) dumped into the hot reservoir.
- Energy balance over a cycle: \(Q_1 = Q_2 + W\), so \(W = Q_1 - Q_2\) and \(Q_1\) is the largest.
- Refrigerator COP: \(\alpha = \dfrac{Q_2}{W} = \dfrac{Q_2}{Q_1-Q_2}\). Heat-pump COP: \(\alpha' = \dfrac{Q_1}{W}\).
- Same machine: \(\alpha_{\text{heatpump}} = \alpha_{\text{fridge}} + 1\).
- Ideal Carnot limits: \(\alpha = \dfrac{T_2}{T_1-T_2}\) (fridge), \(\dfrac{T_1}{T_1-T_2}\) (heat pump), \(T\) in kelvin.
- COP is benefit-per-cost, not efficiency; it can exceed 1. Linked to a Carnot engine by \(\alpha = (1-\eta)/\eta\).
- Clausius statement: heat cannot flow cold → hot as the sole result; a perfect (infinite-COP) refrigerator is impossible.