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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.

Normalization of the reverse law gives an exact finite identity:The sum notation describes discrete paths; the corresponding path integral has the same change-of-variable argument. Jensen's inequality for the exponential function then gives $e^{-⟨{\Sigma }_F{⟩}_F}\le 1$, hence $⟨{\Sigma }_F{⟩}_F\ge 0$. Equivalently this mean is the Kullback-Leibler divergence between the forward path law and the reverse path law pulled back by $\Theta$. Equality requires those two laws to agree. ^[ref]

Rare negative entropy production contributes disproportionately to this expectation through $e^{-{\Sigma }_F}$. Missing such events in a finite sample can spoil an estimate of the identity. Matching probability support is also essential: reverse paths absent from the forward law cannot be recovered by cancelling forward weights. This Second law of thermodynamics statement concerns an ensemble average; an individual negative value is compatible with it.

Integrating the Crooks fluctuation theorem over work and using normalization of the reverse probability distribution givesConsequently Jensen's inequality gives $⟨W{⟩}_F\ge \Delta F$. Some realizations may nevertheless have $W<\Delta F$. This is a work relation with equilibrium endpoint preparations, while the integral fluctuation theorem for total entropy production uses the actual final forward distribution. ^[ref]

No cyclic device can have as its sole effect the extraction of heat from one thermal reservoir and its complete conversion into net work. A putative device absorbing $Q_h>0$ and returning to its starting state with $W=Q_h$ and no other transfer would conserve energy, yet violate the Second law of thermodynamics. Rejecting heat to colder surroundings is an essential part of the ordinary two-thermal reservoir heat engine budget.^[ref]

The cyclic and sole-effect qualifications matter. A single isothermal expansion of an ideal gas can absorb heat and deliver equal work while leaving its volume changed; restoring the initial state must also be counted. Likewise, a device discharging stored internal energy or consuming fuel has another effect. Converting supplied electrical work completely to heat through Joule heating reverses the proposed conversion and is allowed. The statement also allows cyclic devices that exchange heat with one thermal reservoir while consuming net work; it forbids positive net work output with that single heat source as the only external change.

Conversely, a hypothetical complete-conversion device could take $100\mathrm{J}$ from the hot thermal reservoir and supply $100\mathrm{J}$ of work to an ideal refrigerator that extracts $100\mathrm{J}$ from the cold thermal reservoir and delivers $200\mathrm{J}$ to the hot one. At $600\mathrm{K}$ and $300\mathrm{K}$, that refrigerator is itself admissible: its reservoir entropy changes cancel. The composite would consume no external work and move a net $100\mathrm{J}$ from cold to hot, violating the Clausius statement of the second law. Together with the construction there, this proves the equivalence of the two impossibility statements.^[ref]

For the two fixed positive-temperature thermal reservoir hypotheses, Carnot's theorem gives the sharper restriction $\eta \le 1-T_c/T_h<1$. A reversible thermodynamic process reaches that bound in the ideal Carnot cycle; an ordinary device's smaller thermal efficiency reflects its entropy production and measurement boundary. An inefficient observed engine is compatible with the Second law of thermodynamics, rather than evidence that the bound has been experimentally determined.

With the Seebeck coefficient convention $dV=-SdT$ at zero electric current, and the transported-heat convention of Peltier coefficient, the Kelvin thermoelectric relation is$T$ is absolute temperature in kelvins; use consistent material references. This scalar linear response relation assumes local thermal equilibrium and reciprocal transport without time-reversal-breaking fields or order. Onsager reciprocal relations supply the symmetry; conservation of energy alone is insufficient. A magnetic field requires a reversed-field reciprocity comparison.^[ref]

In particular, the A-minus-B thermocouple coefficient in Seebeck effect is $S_A-S_B$, whereas absorbed junction heat for conventional current A to B uses $S_B-S_A$. There is no contradiction: terminal voltage order and crossing direction are separate choices. Multiplying a measured voltage by an absolute temperature without specifying these choices does not determine junction heating.

A short reciprocity calculation makes the factor $T$ visible. At a common reference temperature, write the local linear response with electrical driving field $\mathcal{E}=-\nabla V$ and heat flux $q$ asAt $j=0$, the Seebeck coefficient convention gives $S=L_{12}/(T{L}_{11})$. At $\nabla T=0$, the Peltier coefficient is $q/j=L_{21}/L_{11}$. Thus $L_{21}=L_{12}$ gives $\Pi =TS$. The chosen forces are conjugate to the fluxes in the local entropy production, $j\mathcal{E}/T+q(-\nabla T)/T^2$; arbitrary rescaling of just one force would spoil this argument.

The relation is local. Varying $S(T)$ permits distributed Thomson effect heating, with $\tau =TdS/dT$; the constant-coefficient model below omits it. Endpoint substitutions into a constant-property formula can therefore break its energy accounting.

Three distinct effects spoil indefinite $T\propto B$ cooling: extra entropy outside the spin subsystem, additional magnetic energy scales, and irreversible heat or work input. They can be distinguished with a thermodynamic derivative. For a smooth equilibrium Helmholtz free energy obeying $dF=-SdT-mdB$, a Maxwell relation givesHere $C_B$ is the total heat capacity participating in the stage's thermal equilibrium. The ideal model has $(\partial m/\partial T{)}_B=-C_{B,\mathrm{mag}}/B$. If its spins share temperature with a field-independent lattice and load of combined heat capacity $C_{\mathrm{other}}>0$, thenThe positive slope still means cooling when $dB<0$, but its smaller magnitude gives less cooling per fractional field reduction. For example, a low-temperature Debye model gives a lattice entropy proportional to $T^3$. As that lattice entropy falls, the spins take up the released entropy; $x$ can no longer remain constant. A weak enough coupling to neglect its equilibrium interaction energy can still be strong enough to exchange heat on the sweep timescale.

Interactions between magnetic moments, including exchange interaction and the fields of neighboring magnetic dipole moments, introduce energy scales that do not vanish with the applied $B$. Once they compete with $k_{B}T$ or $\mu B$, the independent-site partition function fails. Correlations and possible magnetic ordering alter the available magnetic entropy and the cooling curve. Simply setting the external magnetic field to zero does not eliminate those interactions. Different refrigerants therefore need their own measured equilibrium properties; there is no universal final temperature from the applied-field ratio alone.

Finally, a real open heat switch has residual thermal conductance; supports, wires and radiation carry incoming heat. Field changes may also dissipate energy. Locally near thermal equilibrium, with incoming heat rate ${\dot{Q}}_{\mathrm{leak}}$ and internal entropy production rate ${\dot{S}}_{\mathrm{prod}}$, the stage balance becomesThis uses a common stage temperature; a non-equilibrated load requires separate balances. The sweep must allow spin-lattice relaxation while remaining fast enough to limit accumulated parasitic heat. At fixed $B$, the same balance explains warming and finite hold time. Useful refrigeration is the controlled absorption of a load's heat during that finite interval, followed by a renewed isothermal process of magnetization to export its entropy.

A single step has negative entropy production with probability $q$, independent of how long an observation lasts. A whole window is negative only if negative steps outnumber positive ones:For even $n$, the tie $K=n/2$ has zero entropy production and is excluded. At $p=3/4$, the negative-window probabilities for $n=1,9,21$ are approximately $0.25$, $0.04893$, and $0.00642$. Longer windows still contain negative increments: their expected number is $nq$. Their total is usually positive because the expected number of positive increments is $np$.

Only odd window lengths are plotted to avoid alternating tie effects. The left panel gives the exact negative-window probability and the upper bound $(2\sqrt{pq}{)}^n$; the right gives expected counts $np$ and $nq$. For steps of fixed duration $\delta t$, physical duration is $n\delta t$. These are counts and probabilities predicted by the model, not bead traces. The bound follows from $\mathrm{Pr}(S_n<0)\le ⟨e^{-S_n/2}⟩=(2\sqrt{pq}{)}^n$. Since $2\sqrt{pq}<1$ for $p\ne q$, negative totals are exponentially suppressed even though the negative-step fraction stays $q$.

Differentiate $E=\rho c^2{a}^3{V}_0$ and use $dE=-PdV$, with $V=a^3{V}_0$. Division by $c^2{a}^3{V}_0$ givesThis is homogeneous adiabatic expansion without entropy production or exchange across the comoving element.

Work in the shock frame, assume a calorically perfect gas with fixed specific-heat ratio $\gamma >1$, and neglect magnetic stresses and body forces across the thin shock. The internal energy per unit mass is $e=p/[(\gamma -1)\rho ]$, and the specific enthalpy is $w=\gamma p/[(\gamma -1)\rho ]$. Integrating the local conservation laws across a thin stationary control volume makes the mass flux $\rho u$, momentum flux $p+\rho u^2$, and energy flux $u(\rho u^2/2+\gamma p/(\gamma -1))$ continuous. Hence the Rankine-Hugoniot conditions for a perfect gas areMultiplication of the last equality by $m$ gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although entropy production occurs on the physical compression branch. The same constant in the polytropic equation of state $p=K{\rho }^{\gamma }$ must therefore not be imposed on both sides. For flow from $z>0$ to $z<0$, both signed velocities are negative; the flux equations and ratios remain valid.

In the Strong-shock Rankine-Hugoniot conditions, $p_1/({\rho }_1{u}_1^2)=1/(\gamma M_1^2)$ tends to zero. Put $r={\rho }_2/{\rho }_1$. The first two flux balances give $u_2=u_1/r$ and $p_2={\rho }_1{u}_1^2(1-1/r)$ to leading order. The energy balance becomesDiscarding the unchanged-state root $r=1$ gives the compressive shock compression ratio and the required limits:

In the shock frame, integrate mass conservation, momentum conservation, and total-energy conservation across a thin interval containing the stationary normal shock wave. No mass, momentum, or energy is stored in the vanishingly thin interval, so each flux has the same value on both sides. For a perfect gas, the specific enthalpy is $h=\gamma p/[(\gamma -1)\rho ]$. The resulting Rankine-Hugoniot conditions for a perfect gas areThe momentum flux includes both transported momentum and pressure force. The energy flux includes kinetic energy and enthalpy, the latter accounting for internal energy and the work needed to push gas through the interval. Entropy production can occur inside the normal shock wave even though total energy is conserved.

For $N$ equal weak shocks, $(1+\delta {)}^N=P_s$, henceAdding the individual increases in specific entropy gives the entropy production in successive weak shocksBy contrast, a single normal shock wave with the same total pressure ratio givesThus a single strong shock produces more entropy than the chain of sufficiently small weak shocks. At fixed $P_s$, the chain approaches zero entropy production as $\delta \to 0$, while the single-shock result stays positive. The gradual compression approaches reversible isentropic flow.

The first part of $\Delta U$ is the change of the quadratic energy associated with the active Ornstein–Uhlenbeck force. The second is minus the change in the particle's potential energy, measured in thermal-noise units. The time integral is the work done by active propulsion, again divided by its noise scale; the full log ratio is the trajectory's time-reversal asymmetry or entropy production.

In a stationary confining state the two endpoint terms remain $O(1)$ as $T\to \infty$ and have zero mean, while the mean active work and the mean log ratio grow proportionally to $T$. Endpoint-term distributions approach time-independent distributions with positive and negative fluctuations. Under mixing assumptions, the time-integrated work has a large-deviation distribution: its central part becomes approximately Gaussian with mean and variance proportional to $T$, while its far tails scale exponentially in $T$. The log-ratio distribution obeys the corresponding fluctuation theorem symmetry.

The first law of thermodynamics, with $\Delta Q$ defined as heat lost by the particle to the bath, isThus $W-\Delta H=\Delta Q$, and part d becomes the local detailed balance identityThe logarithmic path-probability ratio is the bath's entropy production $\Delta S_{\mathrm{bath}}=\Delta Q/T$ in units of $k_B$.

An original deliberately local model makes the cooling quantitative. Fix a total amount $n$ of material, let $x$ be the mole fraction of solid, and stipulate molar entropiesThe constants $a$ and $b$ have units of molar entropy per temperature and molar entropy, respectively. Ignore their pressure dependence and the cell's own heat capacity. This is a local thermodynamics model, not an extrapolation to zero temperature. A constant positive $b$ cannot remain valid once the solid's spins order.

For a reversible thermodynamic process with no heat exchange, total entropy is constant. Differentiating giveswhen $b>aT$ and $x<1$. Define $T_0=b/a$; solving between two states givesIn particular, stipulate $x_0=0$, $T_i=T_0/4$ and $x=1/8$. ThenThe last expression is the local reversible heat requirement per mole frozen at the final state, divided by $b{T}_0$; it is not the heat exchanged during the isolated compression. That compression exchanges no heat. This distinction separates the latent-heat sign from the actual boundary heat budget.

The utility of the model is its explicit condition: $S_s>S_l$ makes $dT/dx<0$, while positive total heat capacity supplies a stable thermal response. More generally, for molar entropies depending only on temperature, let $C_l,C_s$ be the molar heat capacities. Then $dT/dx=-T(S_s-S_l)/[(1-x)C_l+x{C}_s]$. Incoming heat or entropy production adds terms to the balance and can overwhelm the ideal cooling. Neither the numerical ratios above nor the straight entropy lines are measurements of helium-3.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

A steady state does not by itself give every fluctuating observable an entropy production symmetry. If the dimensionless total entropy production ${\Sigma }_t$ has the same-law finite relation and a large deviation principle with rate function $I$ for ${\Sigma }_t/t$, then, on its common domain,The first expression denotes leading exponential scaling; the second follows by taking the scaled logarithm of the probability ratio. A finite-state, autonomous Markov process with time-even states, positive bidirectional rates and a positive stationary distribution supplies a standard setting for a finite total-entropy production relation. The stochastic system entropy is then bounded, so its endpoint contribution does not change the long-time rate function of bath entropy production. ^[ref]

Unbounded state spaces can have endpoint fluctuations on the scale relevant to a large deviation principle. Dropping system entropy, changing the measured observable to heat, or changing the reverse preparation therefore requires a fresh argument. The exact finite integral fluctuation theorem, a same-law finite relation, and an asymptotic rate function symmetry are distinct claims.