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A Carnot cycle realizes the reversible bound in Carnot's theorem. Its four steps for an ideal gas working substance are:

The temperature versus entropy plot in the Carnot's theorem figure is therefore a rectangle traversed clockwise. Because $\delta Q_{\mathrm{rev}}=TdS$, its width is $\Delta S=Q_h/T_h$. For the illustrated engine, $\Delta S=5/3\mathrm{J}{\mathrm{K}}^{-1}$: the upper step absorbs $600\times 5/3=1000\mathrm{J}$, the lower step rejects $300\times 5/3=500\mathrm{J}$, and the enclosed area is the $500\mathrm{J}$ net work. Geometry displays the thermal efficiency without solving the gas motion. The two vertical steps are isentropic processes, not constant-temperature steps. ^{[ref] [ref]}

Reversible heat transfer is a limit of vanishing temperature differences; ordinary finite-rate transfer introduces entropy production. The Carnot cycle is a performance benchmark rather than a guarantee of useful power. Running it backward gives the ideal operation of a refrigerator or heat pump, with work supplied to move heat toward the hotter thermal reservoir. ^[ref]

No cyclic device can have as its sole effect the transfer of heat from a colder thermal reservoir to a hotter thermal reservoir. The word “sole” includes the surroundings: the device must return to its starting state, and there must be no net supplied work, depletion of a stored resource, or accompanying change elsewhere. The statement excludes a self-running refrigerator, while allowing an ordinary refrigerator or heat pump supplied with work.^[ref]

For a proposed transfer $Q>0$ from a fixed cold thermal reservoir at $T_c$ to a fixed hot thermal reservoir at $T_h>T_c>0$, with no other effect, the total entropy change would beThe first law of thermodynamics permits this energy balance; the Second law of thermodynamics forbids the negative total entropy change. For $Q=600\mathrm{J}$, $T_c=300\mathrm{K}$ and $T_h=600\mathrm{K}$, it is $-1\mathrm{J}{\mathrm{K}}^{-1}$. Direct heat transfer in the opposite direction has positive total entropy change. A powered heat pump has a different budget: it supplies work and delivers more heat to the hot side than it extracts from the cold side.

A short equivalence argument links this restriction to the Kelvin-Planck statement of the second law. Couple a hypothetical device that moves $600\mathrm{J}$ from cold to hot with no other effect to the heat engine illustrated above, which absorbs $1000\mathrm{J}$, rejects $600\mathrm{J}$, and delivers $400\mathrm{J}$. The cold-side transfers cancel; the combined cyclic device takes $400\mathrm{J}$ from the hot thermal reservoir and delivers exactly $400\mathrm{J}$ of work, violating the Kelvin-Planck statement of the second law. Matching transfers is an ideal composition argument, not a claim that such a device exists.^[ref]

A heat engine uses a repeating thermodynamic cycle to turn part of an incoming heat transfer into net outgoing work. In the two-thermal reservoir model, the hot thermal reservoir supplies heat $Q_h>0$ at fixed temperature $T_h$, and the cold thermal reservoir receives heat of magnitude $Q_c>0$ at fixed temperature $T_c<T_h$. A Stirling engine makes the repeated motion visible; the abstract energy budget applies independently of its particular mechanism.^[ref]

Choose the entire cyclic device as a closed system, with no net change in bulk kinetic energy or potential energy. Count heat entering as positive and net work delivered as positive, as in the first law of thermodynamics. Thus $Q_h$ enters with a plus sign and $Q_c$, although a positive magnitude, enters with a minus sign. Assume these two heat transfers and the net work are the only transfers across the boundary. Over one complete thermodynamic cycle, all stored energy, including internal energy $U$, returns to its starting value:The device does not accumulate heat from cycle to cycle. More precisely, heat is a transfer, while internal energy is what can be stored; even a component that temporarily stores internal energy must return to its initial state for this boundary to be cyclic.^[ref]

For engine operation, $Q_h>Q_c$ and $W>0$. Its thermal efficiency isUse net work, after subtracting any work supplied during compression or required by auxiliaries included in the chosen device. For periodic operation, average power satisfies $P_{\mathrm{out}}={\dot{Q}}_h-{\dot{Q}}_c$ and $\eta =P_{\mathrm{out}}/{\dot{Q}}_h$, with the rates averaged over complete thermodynamic cycles.^[ref]

For a stipulated model at $T_h=600\mathrm{K}$ and $T_c=300\mathrm{K}$, let $Q_h=1000\mathrm{J}$ and $Q_c=600\mathrm{J}$ per thermodynamic cycle. Then $W=400\mathrm{J}$ and $\eta =2/5=40\%$. At two thermodynamic cycles per second, these become $2000\mathrm{W}$ incoming heat transfer, $1200\mathrm{W}$ outgoing heat transfer, and $800\mathrm{W}$ net power. These are constructed values, not measurements.

Different signs describe different tasks. With the same directions of heat transfer, $Q_h=Q_c$ gives $W=0$: a device transmitting heat without net work output. If $Q_c>Q_h$, net work is supplied to that device. A refrigerator or heat pump instead extracts heat from the cold side and delivers heat to the hot side, using positive input work $W_{\mathrm{in}}$ and the balance $Q_h=Q_c+W_{\mathrm{in}}$. Their useful output is heat transfer, measured by a coefficient of performance. Finally, a one-time isothermal expansion can have $Q=W$ and $\Delta U=0$ while leaving the volume changed: it is not a complete thermodynamic cycle. A warming device also fails the cyclic assumption, and a fuel-consuming device requires an accounting of matter crossing its boundary.

For two fixed thermal reservoirs at positive thermodynamic temperatures, the Second law of thermodynamics strengthens the energy balance toEquality requires the overall operation to be a reversible thermodynamic process, including its heat transfer to the thermal reservoirs. The Carnot cycle explains how the reversible benchmark is constructed, and Carnot's theorem explains its universality and the entropy production penalty. Use kelvin values, never ratios of degree Celsius readings. The example above lies below the $50\%$ bound: its total entropy production is $-1000/600+600/300=1/3\mathrm{J}{\mathrm{K}}^{-1}$ per thermodynamic cycle.

This is useful before building a machine: warmer cooling surroundings reduce the possible thermal efficiency, even if the mechanism is unchanged. A small observed thermal efficiency does not by itself test the Second law of thermodynamics; heat leaks, friction, incomplete recovery of work, and the measurement boundary can all reduce the reported ratio. A meaningful comparison requires the actual thermal reservoir temperatures, matched heat and net work measurements, and the hypotheses of Carnot's theorem. Finite bodies whose temperatures change are treated separately there.

The practical question already appears in the 1824 Réflexions sur la puissance motrice du feu et sur les machines propres à développer cette puissance by Sadi Carnot: pp. 6–8 ask whether machines driven by heat have a fundamental limit and whether changing the working substance can improve their output. Pages 9–11 describe heating and cooling in real machines, but use the historical caloric theory, rather than the modern first law of thermodynamics balance. The historical discussion under Carnot's theorem explains that distinction.

A heat pump uses work to deliver heat to the region we want to keep warm. Its useful output is the hot-side delivery $Q_h$. The same operating machine can be described as a refrigerator when the cold-side extraction $Q_c$ is the objective: which output is useful changes the coefficient of performance, while the energy balance stays $Q_h=Q_c+W$. Heating a water bath by taking heat from a colder water bath makes this distinction visible in the heat-pump calorimetry experiment below. ^[ref]

In the illustrative budget from refrigerator, supplying $1\mathrm{kJ}$ of work delivers $4\mathrm{kJ}$ of useful heat, giving a heating coefficient of performance of $4$. The additional $3\mathrm{kJ}$ came from the colder region. Compare an ideal electrical heater whose entire input work becomes heat in the target region: it delivers $1\mathrm{kJ}$ for that input. The appeal of a heat pump is this ability to move existing energy, while the Second law of thermodynamics limits the work saving for a given pair of temperatures.

A motivating bench experiment asks whether a hot bath receives more heat than the electrical work supplied to a heat pump. PHYWE's authored operating instructions, section 4, describe a sealed vapor-compression refrigeration apparatus with its cold and hot coils immersed in separate water vessels. The quantitative procedure calls for insulated vessels, stirring to reduce temperature gradients, timed temperature readings, and a wattmeter for the input. It waits about five minutes for operating conditions to settle before measurements and explicitly includes the glass vessel and immersed copper pipe when calculating the heated body's heat capacity. One bath cooling while the other warms makes the origin of the delivered energy observable. ^[ref]

The following analysis extends that procedure to simultaneous measurements of both finite baths. Weigh the water in each vessel; determine the effective heat capacities $C_c$ and $C_h$, including immersed hardware, by a separate known electrical-input calibration over the working temperature range. A matched pair of calibrated thermometers placed in stirred water measures bulk $T_c(t)$ and $T_h(t)$; keeping probes away from the coils avoids mistaking a local contact temperature for the whole bath. Record the room temperature and synchronise the electrical meter and thermometry. Use short intervals after startup, with no boiling or freezing and small enough temperature changes that constant effective heat capacities are a reasonable approximation. These are proposed measurement controls, rather than reported results from the manual.

Over a common interval $[t_0,t_1]$, the uncorrected bath estimates and measured electrical input areHere the wattmeter records real electrical power $P_{\mathrm{el}}$. The integral handles changing input, and the signs make both estimated useful heat quantities positive during heating and cooling. The effective heat capacity includes water mass times its specific heat capacity plus the response of the vessel and immersed hardware. Without that correction, equal temperature changes need not imply equal heat transfers.

Insulation reduces heat leaks but does not establish an isolated system. A warmer bath can lose heat to the room while a colder bath gains it. As a proposed control, measure pump-off drift at matched bath and room temperatures, with the same stirring and insulation. Writing $L_h>0$ for outward hot-bath leak power and $L_c>0$ for inward cold-bath leak power, the bath balances giveIntegrate these rates over the same interval before dividing by $W_{\mathrm{el}}$. For example, in a pump-off hot-bath control, $L_h=-C_{h}d{T}_h/dt$; in the cold-bath control, $L_c=C_{c}d{T}_c/dt$. Check that a stopped machine does not create a different leak path from the running one; otherwise use an independent bath calibration and keep that discrepancy as an uncertainty. Repeat at the same starting temperatures to distinguish drift from a change in COP as the baths separate.

In the illustrated toy model, $C_h=C_c=4500\mathrm{J}{\mathrm{K}}^{-1}$, $P_{\mathrm{el}}=50\mathrm{W}$, ${\dot{Q}}_h=150\mathrm{W}$, ${\dot{Q}}_c=100\mathrm{W}$, $L_h=15\mathrm{W}$, and $L_c=5\mathrm{W}$ are stipulated constants for $180\mathrm{s}$. The slopes are $d{T}_h/dt=0.030\mathrm{K}{\mathrm{s}}^{-1}$ and $d{T}_c/dt=-19/900\mathrm{K}{\mathrm{s}}^{-1}$. Consequently,The raw heating and cooling ratios are $2.7$ and $1.9$, whose difference is $0.8$. Correcting the bath leaks gives $3$ and $2$, whose difference is one. The apparent failure of $Q_h=Q_c+W$ was incomplete measurement of heat, not a violation of the first law of thermodynamics.

Real apparatus can also release motor heat and pipe heat directly to the room, and store internal energy during startup. For a machine boundary that receives electrical work, an additional outward heat transfer $Q_{\mathrm{room}}$ gives $Q_h+Q_{\mathrm{room}}=Q_c+W_{\mathrm{el}}-\Delta U_{\mathrm{machine}}$. The residual $Q_h-Q_c-W_{\mathrm{el}}$ therefore tests the boundary and corrections; do not force it to zero. Include any auxiliary electrical work in the input if quoting whole-apparatus COP, and account separately for stirring work deposited in a bath. Ignoring vessel heat capacity, sensor lag or room leaks can bias the inferred result. These limitations follow from extending the measured energy budget, and the manual's explicit hardware correction illustrates why such boundaries matter.

Finally, these vessels are finite bodies whose temperatures change; they are not fixed thermal reservoirs. Their ratio of integrated useful heat to integrated input work is an interval COP. A Carnot's theorem comparison using simultaneous bath temperatures is a local approximation only when the machine settles much faster than the baths change. An exact reversible comparison for changing bath temperatures instead uses the bath entropy changes, for example $C_h\ln (T_{h,1}/T_{h,0})+C_c\ln (T_{c,1}/T_{c,0})$ for constant heat capacities, negligible net machine entropy change and no room leaks. The experiment's beauty is that two thermometers, known heat capacities and an electrical-energy measurement expose both conservation of energy and the practical gap from reversibility.

A reversible heat pump maintaining $T_h$ against an external $T_c<T_h$ has heating coefficient of performance $T_h/(T_h-T_c)$. If its maximum work input is ${\dot{W}}_{\max }$ and the building loses heat at rate $\lambda (T_h-T_c)$, maintaining equilibrium requires $\lambda (T_h-T_c{)}^2\le {\dot{W}}_{\max }{T}_h$. This yields the displayed minimum external temperature, provided it lies in the physical positive-temperature range. If the rating instead specifies maximum delivered heat, the bound is $T_h-T_c\le {\dot{Q}}_{h,\max }/\lambda$; a heat-output rating cannot be substituted for a work-input rating in the Carnot formula.

A heat pump delivers $Q_H=Q_S+W$: electrical work moves heat from a source as well as contributing energy itself. Its heating coefficient of performance can therefore exceed one, whereas direct resistance heating supplies at most its electrical input as useful room heat. For an ideal reversible pump,with absolute temperatures and $T_S<T_H$. This is an ideal limit; actual costs also depend on compressor performance and tariffs.

For the annual ground-temperature wave, model the soil as a homogeneous semi-infinite conductor, with depth positive downward and a prescribed surface temperature. Here $\kappa$ is thermal conductivity and $C$ is heat capacity per unit volume, so the thermal diffusivity is $\kappa /C$. Conservation of energy gives $C{\partial }_{t}T=\kappa {\partial }_z^{2}T$. After transients decay, put $T-T_0=\mathrm{Re}[A{e}^{i\omega t-qz}]$. Substitution gives $q^2=i\omega C/\kappa$; the solution bounded at depth uses $q=(1+i)/\delta$. Matching the surface then givesThis approximation identifies the ground surface with air temperature and neglects the pump's perturbation of the ground, as required by the model. The attenuation and phase delay arise from heat diffusion, not propagation at one fixed wave speed.

The graph below covers a full annual cycle. At $z/\delta =\pi /2$, the amplitude is $20e^{-\pi /2}=4.16$ K and the minimum lags January by a quarter-year. At $z/\delta =\pi$, the amplitude is only $0.864$ K and the minimum lags by half a year: the ground is slightly warmer than its annual mean in January.

The seasonal source depth for a reversible heat pump has a subtle limit: a deep source approaches $T_0$ and avoids severe winter cooling, but infinite depth is not the thermal optimum of this single-frequency model. In January, with $x=z/\delta$, maximize $T_S=T_0-T_A{e}^{-x}\cos x$. The first and largest maximum is at $x=3\pi /4$, where $\cos x+\sin x=0$, giving $z=3\pi \delta /4$ and $T_S\simeq 16.34^{\circ }\mathrm{C}$. It is also the optimum of the ideal annually integrated heating work: the heating load is even about January, so its weighted sine term integrates to zero and the depth dependence is again proportional to $e^{-x}\cos x$. Installation cost, finite tube spacing, heat extraction, weather and real pump performance alter this result. The supplied model therefore supports a finite seasonal thermal optimum, while not fixing a practical burial depth in metres.

Use absolute temperatures $T_c=278.15\mathrm{K}$ and $T_h=298.15\mathrm{K}$. Let ${\dot{Q}}_c$ be the heat extracted from the ground, ${\dot{Q}}_h$ the heat delivered indoors, and $P$ the electrical input power. For a reversible heat pump, the entropy transferred per unit time balances:Consequently the coefficient of performance isThe requested heat delivery is a rate, since the duration is unspecified:Of this, $13.9\mathrm{kW}$ comes from the ground and $1.0\mathrm{kW}$ from the electricity. A coefficient of performance greater than one is consistent with conservation of energy because the heat pump transports heat as well as converting electrical energy.

Two equivalent second-law statements. The Clausius statement of the second law forbids a cyclic device whose sole net effect is transferring heat from a colder reservoir to a hotter one. A refrigerator can make that transfer when supplied with work, so the sole-effect qualification matters. The Kelvin-Planck statement of the second law forbids a cyclic device whose sole net effect is drawing heat from a single reservoir and converting all of it into work. A heat engine can produce work when it also rejects heat to a colder reservoir.

The two statements are equivalent by combining a supposed violating device with an ordinary reversible engine or refrigerator: one can cancel one reservoir exchange and obtain a violation of the other statement. The following original schematic shows the forbidden sole effects, with arrows indicating energy flow.

Microscopic fluctuations. The second law is a macroscopic statistical law. A finite fluctuation can transfer a small heat amount $Q$ from cold to hot, with reservoir entropy change $Q/T_h-Q/T_c<0$. Such an individual event does not imply that a cyclic engine can extract systematic useful work from equilibrium. Entropy-decreasing trajectories become exceedingly improbable when the entropy involved is large compared with $k_B$. Under suitable microscopic reversibility conditions, fluctuation relations quantify the relative probabilities of positive and negative entropy production and give nonnegative mean total entropy production. Thus a deterministic no-fluctuation reading has a limit at small scales or short times; the statistical second law remains applicable.

Maximum heating rate. Interpret the quoted $0.5\mathrm{kW}$ as the maximum work-input rate ${\dot{W}}_{\max }$ to an ideal reversible heat pump. This interpretation is needed to obtain a heating rate larger than its electrical input; the wording does not explicitly distinguish input power from delivered heat. A reversible cycle satisfiesTherefore the heating coefficient of performance isWith $T_h=293\mathrm{K}$ and $T_c=273\mathrm{K}$,The additional heat comes from the outdoor reservoir; it is not produced by converting work into more energy than was supplied.

Lowest outdoor temperature. Let $\Delta T=T_h-T_c>0$. The largest available heating rate is ${\dot{W}}_{\max }{T}_h/\Delta T$, while the house loses $\lambda \Delta T$. Equilibrium can be maintained whenAt the limiting outdoor temperature,HenceThis is the indoor-temperature limit of a Carnot heat pump. It assumes the ideal pump retains its stated input-power capability over the operating range. Real heat exchangers require finite temperature differences and incur irreversibility, so the reversible result is an ideal upper performance bound.

Conventional heater. At this limiting outdoor temperature, the required delivered heating power is simply the house's heat loss:This is heat delivered to the house. The fuel-input power would be $3.83/\eta \mathrm{kW}$ for heater efficiency $\eta$. At the earlier outdoor temperature of $273\mathrm{K}$, the required delivered heat would instead be $0.1(293-273)=2.00\mathrm{kW}$; the final question's reference temperature must be kept clear.

If $0.5\mathrm{kW}$ is read literally as a maximum delivered heat rate rather than a work-input rating, the alternative answers are ${\dot{Q}}_{h,\max }=0.5\mathrm{kW}$, $T_{c,\min }=293-0.5/0.1=288\mathrm{K}$, and a conventional heater delivering $0.5\mathrm{kW}$ at that limit. Such a pump could not balance the stated house loss at $273\mathrm{K}$. Both readings follow consistently once the ambiguous power rating is defined; the Carnot calculation above uses the work-input reading.

Electricity is a strong early decarbonization target because large stationary sources are comparatively accessible and a low-carbon grid enables cleaner vehicles, heat pumps and industry. For example a heat pump with coefficient of performance three divides electricity-related heat emissions by three per unit useful heat. But grid decarbonization does not automatically outrank every efficiency, methane, land-use or industrial intervention. Marginal abatement cost, cumulative emissions, construction time and available technology determine priorities. A very coal-intensive grid makes electrification alone less effective; a clean grid does not remove aviation or process emissions. The statement is a plausible strategic emphasis, not a deduction valid for every country and timescale.

The Peltier effect is heat transfer associated with an electric current crossing a junction between materials with different Peltier coefficients. At fixed junction temperature, its contribution changes sign when the conventional current reverses. A junction can absorb heat from its surroundings for one direction and release it for the other. Joule heating, proportional to the square of the electric current, accompanies this transport but does not reverse sign. Consequently, observing a warm device alone does not identify the Peltier effect.

This makes the Peltier effect useful as an electrically controlled heat pump: reversing the supply can switch an object's thermal role without changing any plumbing. In thermoelectric cooling, however, the useful cooling is the junction's absorbed heat minus returning heat conduction and its share of Joule heating. Increasing electric current eventually makes cooling worse. The distinction between transported heat and net cooling is the central practical lesson of the thermoelectric cooler energy balance.

Vapor-compression refrigeration circulates a working fluid through cold and hot coils. In the cold coil, liquid becomes gas while absorbing heat from the refrigerated region. Compression supplies work and raises the pressure and temperature; the hot coil releases heat as the gas returns to liquid. A restriction reduces the pressure before the next cold-side passage. Changing the pressure changes the temperature of the phase transition, letting one circulating fluid receive and release heat at different temperatures. This physical implementation serves both refrigerators and heat pumps; its ordinary dissipative operation differs from the reversible Carnot cycle used to bound performance. ^{[ref] [ref]}