Refrigerators, heat pumps & the second law

A heat engine takes Q_H from a hot reservoir, delivers work W and rejects Q_C to a cold one, tracing its cycle clockwise on the p-V diagram. Reverse the order of the strokes and the loop runs anticlockwise: the enclosed area is now work done on the gas, the low-temperature stroke absorbs heat and the high-temperature stroke rejects it. Declare the three quantities as positive magnitudes and the first law over one cycle, where ΔU = 0, reads Q_H = Q_C + W. A refrigerator that removes 150 J from its cabinet per cycle while its compressor does 50 J of work must dump 200 J through the coils on its back: not 150 J, and never less than it took in. The same device is a refrigerator if the cold reservoir is the object of interest and a heat pump if the hot one is. The hardware and the energy ledger are identical; only the customer's question changes.

Efficiency divides useful output by the energy consumed to make it, and for an engine it is bounded by 1. A refrigerator has no such ratio: the heat it removes is not made from the work, it is moved by it. So its figure of merit is the coefficient of performance, K_cool = Q_C/W, and for the same device used as a heater Kₕₑₐₜ = Q_H/W = (Q_C + W)/W = K_cool + 1. For the 150 J per 50 J refrigerator above, K_cool = 3.0 and Kₕₑₐₜ = 4.0. Neither number is a percentage and neither is capped at 1: a heat pump with Kₕₑₐₜ = 4 delivers 4 kW of heat for 1 kW of electricity, three of them lifted from outside. The unit check is trivially J/J, so the sign convention check carries the weight. Both coefficients are ratios of positive magnitudes, and a student who carries the signed work done by the gas, W = −50 J, into them gets a negative coefficient that means nothing.

For any reversible cycle between two reservoirs the heats exchanged stand in the ratio of the temperatures, Q_H/Q_C = T_H/T_C. Running the Carnot cycle backwards leaves it untouched, since each isothermal heat is nRT ln(V₂/V₁) with the same volume ratio either way. Substitute into the definitions: K_cool, max = Q_C/(Q_H − Q_C) = T_C/(T_H − T_C) and Kₕₑₐₜ, max = T_H/(T_H − T_C). A refrigerator cabinet at 275 K in a 295 K kitchen has a ceiling of 275/20 = 13.75; a freezer at 255 K in the same kitchen has 255/40 = 6.4, less than half, for doubling the lift. A heat pump warming a house at 293 K has Kₕₑₐₜ, max = 293/20 = 14.65 with outdoor air at 273 K but only 293/40 = 7.3 at 253 K, which is why its output sags on the coldest nights. Check the limits: as T_C → T_H the bound diverges, since moving heat across no temperature difference costs no work; as T_C → 0 it falls to zero, so each joule taken from a body near absolute zero costs unbounded work. And Kₕₑₐₜ, max = 295/20 = 14.75 for the fridge case is exactly 1/(1 − 275/295), the reversed engine seen from its other side.

The Clausius statement: no device operating in a cycle can have as its sole effect the transfer of heat from a colder body to a hotter one. The Kelvin-Planck statement: no device operating in a cycle can have as its sole effect the absorption of heat from a single reservoir and the performance of an equal amount of work. Every word is load-bearing. A refrigerator does move heat from cold to hot, but not as its sole effect: it consumes work and rejects more heat than it collected. An ideal gas expanding isothermally does convert heat entirely into work, but it does not operate in a cycle: it ends expanded, and returning it to its start undoes the conversion. Neither statement mentions an amount of energy; both forbid a process the first law would permit. Clausius names the fact that a refrigerator needs W > 0 and hence that K_cool is finite; Kelvin-Planck names the fact that an engine needs a cold reservoir and hence that η < 1.

Suppose a Clausius violator moves 1000 J from a reservoir at 300 K to one at 400 K with no work. Run a Carnot engine between the same reservoirs, efficiency 1 − 300/400 = 0.25, fed with that 1000 J: it delivers 250 J of work and rejects 750 J to the cold reservoir. The hot reservoir receives 1000 J and gives up 1000 J, so it is untouched. The cold reservoir gives 1000 J and gets 750 J back, a net loss of 250 J. The composite has taken 250 J from a single reservoir and turned all of it into work: a Kelvin-Planck violator. Now the other way. Suppose a Kelvin-Planck violator draws 1000 J from the 400 K reservoir and delivers 1000 J of work. Feed that work into a Carnot refrigerator, K_cool = 300/100 = 3: it lifts 3000 J from the cold reservoir and rejects 4000 J to the hot one. Net: the hot reservoir loses 1000 J and gains 4000 J, up 3000 J; the cold reservoir is down 3000 J; no work crosses the boundary. That is 3000 J moved from cold to hot as the sole effect, a Clausius violator. Each statement implies the other, so they are one law written from two sides of the same cycle.

The first law is an equation: energy in equals energy out plus energy stored, and it is satisfied by a refrigerator with W = 0 just as well as by a real one. The second law is an inequality, and it never says how much energy there is; it says which of two energy-conserving processes may happen. Heat may flow from hot to cold unaided, and must not flow the other way unaided. Work may be converted wholly into heat at any temperature, and heat must not be converted wholly into work in a cycle. A coefficient of performance may sit anywhere below T_C/(T_H − T_C), and must not sit above it. Three features follow. The law forbids but never compels: nothing obliges the heat to flow at all. It sets no rate: a perfectly insulated cabinet holds its temperature with the compressor off, and the rate at which a real refrigerator must remove heat is fixed by its leaks, not by the second law. And every allowed process spends something the reverse cannot recover, an asymmetry the next topic turns into a state function called entropy.