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No cyclic heat engine exchanging heat only with two fixed thermal reservoirs at positive absolute temperatures $T_h>T_c$ can exceed the thermal efficiency of an engine whose cycle is a reversible thermodynamic process between them. All such reversible engines have the same thermal efficiency, regardless of their working substance. Let $Q_h>0$ be heat absorbed from the hot thermal reservoir and $Q_c>0$ the magnitude rejected to the cold one. ThenThe work convention is the one in the first law of thermodynamics: $W>0$ is net output by the engine. Use absolute thermodynamic temperatures, such as values in kelvin; ratios of readings in degree Celsius are invalid because their zero is shifted.

For $T_h=600\mathrm{K}$, $T_c=300\mathrm{K}$ and $Q_h=1000\mathrm{J}$, the maximum thermal efficiency is $50\%$. A reversible engine delivers $W=500\mathrm{J}$ and rejects $Q_c=500\mathrm{J}$. The thermal reservoirs have entropy changes $-1000/600$ and $+500/300\mathrm{J}{\mathrm{K}}^{-1}$, which cancel. The engine's own entropy change is zero over its thermodynamic cycle.

A short derivation now becomes transparent: the Second law of thermodynamics gives $-Q_h/T_h+Q_c/T_c\ge 0$, and the first law of thermodynamics gives $W=Q_h-Q_c$. Together they yield the bound and, more precisely,At fixed $Q_h,T_h,T_c$, every extra unit of total entropy production reduces output by $T_c$ units of energy. ^[ref]

Fixed thermal reservoirs are essential. Finite bodies cool and warm, so their initial temperature ratio cannot be used throughout an entire extraction process. As a computed contrast, the bodies with equal constant heat capacity in the entropy example would reach $T_f=\sqrt{600\times 300}=424.26\mathrm{K}$ if a reversible engine extracted work until their temperatures matched. Zero total entropy change gives this geometric mean; the first law of thermodynamics then gives $W_{\max }=C(600+300-2T_f)=5147.19\mathrm{J}$. Direct contact reaches $450\mathrm{K}$ and extracts no work.

Carnot's 1824 book asked whether the useful output of machines driven by heat was fundamentally limited and whether changing the working substance could improve that limit. Its reversible comparisons are the historical setting of Carnot's theorem. Carnot used the then-current caloric picture, treating heat as conserved in its passage through an engine; the modern balance $Q_h-Q_c=W$ should not be read back into that text.^[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]

For the cyclic two-side energy balance in refrigerator, define the dimensionless cooling and heating coefficients byThis identity applies even to irreversible machines, provided both coefficients refer to the same operating point, work input and boundary with no additional energy transfer or storage. It does not compare two separately operated appliances. Unlike thermal efficiency of a heat engine, a COP can exceed one: its numerator counts heat moved, including energy taken from another region. ^[ref]

To obtain temperature-only bounds, impose stronger assumptions: a cyclic machine exchanges heat only with two ideal thermal reservoirs maintained at uniform, fixed absolute thermodynamic temperatures $0<T_c<T_h$, and receives work through an ideal input carrying no entropy. Include the machine and the heat transfer interfaces in the entropy production accounting. Because its own entropy returns to its initial value each cycle, the Second law of thermodynamics givesCombining this inequality with $Q_h=Q_c+W$ givesEquality requires a reversible thermodynamic process for the entire cycle and its contacts, as in a reversed Carnot cycle. These are Carnot's theorem bounds independent of the working substance. Use kelvin values, not degree Celsius readings, in the numerators; the temperature difference can be expressed in either unit. ^{[ref] [ref]}

The diagram fixes $T_c=280\mathrm{K}$ and sets $T_h=320\mathrm{K}$, so the reversible bounds are $280/40=7$ and $320/40=8$. The stipulated machine with ${\mathrm{COP}}_{\mathrm{r}}=3$ and ${\mathrm{COP}}_{\mathrm{heat}}=4$ satisfies both. Its reservoir entropy production per cycle is $4000/320-3000/280=25/14\mathrm{J}{\mathrm{K}}^{-1}>0$. Increasing the temperature lift $\Delta T=T_h-T_c$ at fixed $T_c$ lowers both bounds. Equivalently, delivering fixed $Q_h$ needs at leastThese expressions reveal why a small temperature lift is attractive. Their divergence as $T_h-T_c\to 0^{+}$ is an ideal reversible limit, not a promise of infinite heating or cooling power. For finite heat transfer rates, the working fluid must be colder than the cold bath and hotter than the hot bath; these contact differences produce entropy and increase the internal temperature lift. Dissipative compression and the pressure-reducing restriction of vapor-compression refrigeration supply further departures from reversibility. Thus bath temperatures set an outer bound, not a prediction of real COP. ^[ref]

At fixed temperature and pressure, a closed reacting system reaches thermochemical equilibrium by minimizing Gibbs free energy over nonnegative species amounts subject to conserved elemental abundances. For an allowed stoichiometric vector $\nu$, an interior minimum has ${\sum }_i{\nu }_i{\mu }_i=0$. More generally the chemical potential vector lies in the row span of the elemental conservation matrix at positive concentrations, with the appropriate one-sided inequalities for absent species. The first law of thermodynamics gives the thermodynamic differential, while the Second law of thermodynamics supplies the minimization direction.

The 2002 experiment trapped a $6.3\mu \mathrm{m}$ bead in water with a $980\mathrm{nm}$ laser, a $100\times$, numerical aperture $1.3$ objective and a piezo-driven microscope stage. A quadrant photodiode tracked bead position at $1\mathrm{kHz}$ with approximately $500\mathrm{Hz}$ bandwidth and reported $15\mathrm{nm}$ resolution. Stage translation drove water past the fixed optical trap at $1.25\mu m/s$. Weak trapping prolonged the transient; stationary recording before translation established the position reference. Stokes drag law calibration was checked against the bead's measured diffusion coefficient. ^[ref]

From calibrated displacement and calibrated harmonic optical trap force, the experimenters evaluated the dimensionless driven-bead dissipation functionalAmong 540 trajectories, negative values became less frequent with duration. Figure 2 compared the negative/positive count ratio with $⟨e^{-{\Sigma }_t^{\mathrm{drive}}}{⟩}_{{\Sigma }_t^{\mathrm{drive}}>0}$; its agreement tested the paper's integrated transient relation. The detailed log-ratio plot in Figure 4 used simulated particle sample trajectories, not bead measurements. Near-zero sign discrimination and stage-start transients limited the earliest points. ^[ref]

For any continuous probability distribution satisfying the same-law relation, integration over $s>0$ yieldsThis conditional relation explains the comparison without equating it to the unconditional integral fluctuation theorem. The experiment supports a specified driven-system relation; it does not establish a universal same-law symmetry for any definition of heat or entropy production. In particular, a transient work-like functional should not silently acquire the actual-distribution endpoint term in finite trajectory entropy production. Negative realizations are compatible with positive mean dissipation and the Second law of thermodynamics; they do not provide a cyclic machine with reliable work output from a single thermal reservoir.

Entropy is an extensive quantity and a state function of an equilibrium system, measured in $\mathrm{J}{\mathrm{K}}^{-1}$. Differences between equilibrium states are calculated along a reversible thermodynamic process:Here heat entering the system is positive and $T$ is its absolute thermodynamic temperature. The reversible path used for calculation need not be the path actually taken. Dividing the actual heat by a single temperature generally fails for an irreversible process.

For example, put two rigid bodies, each with constant heat capacity $C=100\mathrm{J}{\mathrm{K}}^{-1}$, at $600\mathrm{K}$ and $300\mathrm{K}$ in an otherwise isolated system. Assume each body has a uniform internal temperature, no phase transition occurs, and no work is extracted. Their final thermal equilibrium is at $450\mathrm{K}$, because the internal energy lost by one equals that gained by the other. Using entropy from heat capacity,so $\Delta S_{\mathrm{tot}}=C\log (9/8)=11.78\mathrm{J}{\mathrm{K}}^{-1}$. One body's entropy decreases while the total increases. ^[ref]

In statistical mechanics, Boltzmann's entropy formula counts equally likely microstates compatible with a macrostate; Gibbs entropy handles a specified probability distribution over microstates. If the number of equally likely microstates doubles, entropy increases by $k_B\log 2$. The logarithm makes multiplicative counts additive for independent systems. These are precise statements about states and probabilities, not a universal measure of visible messiness. Calling entropy merely "disorder" obscures the thermal calculation above and hides which microstates are being counted. ^[ref]

The usefulness of entropy is that measurable heat capacities determine its changes, and the Second law of thermodynamics turns those changes into constraints on processes. It also connects macroscopic measurements to microscopic descriptions without requiring us to track every particle.

For the insulated gas-only free expansion of an ideal gas, the equilibrium endpoints are $(T,V_1)$ and $(T,V_2)$ with $V_2>V_1$. Because entropy is a state function, calculate their difference along a hypothetical reversible thermodynamic process: an isothermal expansion using a frictionless piston and a thermal reservoir at $T$. This comparison process receives heat and delivers work, even though the actual free expansion does neither. Its reversible heat increment is $\delta Q_{\mathrm{rev}}=pdV=nRTdV/V$, soHere $R$ is the molar gas constant, and the logarithm acts on a dimensionless ratio. The reversible reference gives the correct endpoint entropy difference because entropy is a state function; putting the actual $Q=0$ into $\Delta S=Q/T$ would misuse a relation requiring reversible heat transfer. For the actual isolated gas, all this increase is entropy production. Thus an adiabatic process can increase entropy and need not be an isentropic process. ^[ref]

In the figure's volume doubling, the ideal gas law gives $p_1=99.768\mathrm{kPa}$ and $p_2=49.884\mathrm{kPa}$, while $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. The reference process receives and delivers $nRT\ln 2\simeq 1728.85\mathrm{J}$ as heat and work; its thermal reservoir loses the same entropy that the gas gains. The actual free expansion has $Q_{\mathrm{in}}=W_{\mathrm{out}}=0$ and produces that entropy internally. Reversibly compressing the gas back at $300\mathrm{K}$ would require $1728.85\mathrm{J}$ of work input and rejection of equal heat. Restoring the gas alone therefore changes the surroundings: the actual release is irreversible.

For a real gas the final temperature may differ, so the ideal-gas logarithm formula cannot simply be copied. Evaluate $S(T_2,V_2)-S(T_1,V_1)$ using that gas's equilibrium properties or a suitable reversible comparison path. If the walls and water exchange heat with it, the Second law of thermodynamics constrains the total entropy of the isolated apparatus; the gas alone is then not an isolated system.

For a closed system with fixed composition in an environment at temperature $T_R>0$ and pressure $p_R$, define $A=U+p_{R}V-T_{R}S$, where $U$ is internal energy, $V$ is volume and $S$ is entropy. The exergy is this availability relative to the equilibrium reference state, $A-A_{\mathrm{eq}}$. When the combined system and environment are isolated except for the specified exchanges, $\Delta S_{\mathrm{tot}}=-\Delta A/T_R$, so the Second law of thermodynamics makes $A$ decrease toward its constrained minimum. Away from equilibrium, the environmental $T_R,p_R$ must not be replaced by the system's instantaneous values.

Choose a positive constant $\Sigma$ and the illustrative interface lawHere $T_h$ and $T_c$ are the hot and cold interface-side temperatures, not distant thermometer readings. This is a low-temperature phonon model under restrictive assumptions, not a universal fit to Kapitza's measured exponent or a material-property table. Its coefficient $\Sigma$ is not the Stefan-Boltzmann constant: the carriers in this model are phonons, rather than photons crossing a vacuum.

The fourth power has a simple counting explanation. In the three-dimensional low-temperature Debye model, a linear phonon dispersion relation gives a density of states proportional to ${\omega }^2$. A phonon has energy $\hslash \omega$ and Bose-Einstein distribution occupation $[\exp (\hslash \omega /k_{B}T)-1{]}^{-1}$. If propagation speeds and the angle-averaged transmission introduce no additional frequency scale over the occupied modes, the outgoing energy flux has the formExtending the upper integration limit assumes occupied modes lie well below the cutoff. Balancing the two outgoing fluxes, with coefficients satisfying zero net heat flux at equal temperatures, gives the chosen difference of fourth powers. Frequency-dependent transmission, different excitations or additional transport channels can change this conclusion.

The finite-drop area-specific thermal resistance follows by factoring a difference of powers:Near thermal equilibrium, write $T_h=T+\delta T$ and $T_c=T$. Expansion gives $q=4\Sigma T^3\delta T+O((\delta T{)}^2)$, so the differential area-specific Kapitza resistance isAt a finite bias, varying only the hot side gives $(\partial q/\partial T_h{)}^{-1}=1/(4\Sigma T_h^3)$; this is generally different from $r_{K,\sec }$. Stating what is held fixed makes a differential resistance reproducible.

For an original normalized example, choose reference scales $T_0>0$ and $A_0>0$, and setFor contact area $A=2A_0$, the total heat transfer rate is $\dot{Q}=2.1472A_0\Sigma T_0^4$. The finite-drop resistance satisfies $\Sigma T_0^3{r}_{K,\sec }=0.2/1.0736=125/671\approx 0.18629$. Using the equilibrium tangent at $T_0$ instead would predict $j=4(0.2)=0.8$, illustrating why a substantial temperature rise requires the nonlinear law. Doubling the common equilibrium temperature reduces $R_K$ by eight in this chosen model.

A series check can be done without guessing any material properties. Stipulate a hot-side slab with dimensionless areal bulk resistance $b_s=\Sigma T_0^3{L}_s/k_s=0.10$, and neglect the cold-side bulk drop. At the same $j=1.0736$, the hot reservoir must have $T_H/T_0=1.2+b_{s}j=1.30736$. The imposed reservoir drop splits into $0.10736$ in the bulk and $0.20$ at the interface. Applying the fourth-power law directly to $T_H$ would assign the bulk drop to the interface and overestimate the heat flux.

Finally, transfer heat at rate $Aq$ between maintained interface-side temperatures $T_h>T_c>0$. The hot side loses entropy at rate $Aq/T_h$, and the cold side gains it at rate $Aq/T_c$. With no interface entropy accumulation, the entropy production per unit area isFor the normalized example it is ${\dot{S}}_{\mathrm{prod}}/(A\Sigma T_0^3)=1.0736(1-1/1.2)=0.178933\dots$. The same energy crosses both sides in a steady state, while total entropy increases, consistently with the Second law of thermodynamics. A finite temperature jump during heat transfer therefore expresses dissipation, not a failure of energy conservation.

The equilibrium calculation determines where the two temperatures end, without determining how long they take. Stipulate a constant thermal conductance $G>0$ and the linear heat transfer law underlying Newton's law of cooling. With $Q$ positive from body 1 to bath 2, the ordinary differential equations areThe dot denotes a derivative with respect to time. This model assumes internal equilibration is fast compared with exchange between the bodies. Adding the two energy rates gives zero; subtracting the temperature rates givesThe temperature difference has exponential decay with time constant $\tau$. For the finite heat bath example and the stipulated $G=2\mathrm{W}{\mathrm{K}}^{-1}=2\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{K}}^{-1}$, $\tau =45\mathrm{s}$, and solving the ordinary differential equations givesAt one time constant, $T_1\simeq 315.865\mathrm{K}$, $T_2\simeq 293.793\mathrm{K}$ and $Q\simeq 3413.451\mathrm{J}$. Equal temperatures are approached as a limit at infinite time. The finite bath causes the temperature difference to shrink faster than the fixed-bath model, because both bodies change temperature.

For an ideal heat bath with the same $G$, set $T_2=T_b=290\mathrm{K}$ rather than evolving its temperature. Newton's law of cooling then yieldsChanging $G$ changes these time constants, while the final temperature is fixed by the capacities and initial conditions. The dynamical model independently checks the Second law of thermodynamics:Its integral over the relaxation reproduces the endpoint entropy change. The useful separation is between capacity, which governs how much the bath temperature shifts, and thermal conductance, which governs the exchange rate. Neither alone establishes a reliable heat bath.

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.

The Helmholtz free energy iswhere $U$ is internal energy, also denoted $E$ elsewhere here, $T$ is absolute temperature, and $S$ is entropy. For the closed system of fixed composition and only pressure-volume work described above, spontaneous relaxation at fixed temperature and volume cannot increase $F$. Stable thermodynamic equilibrium is a global minimum over the permitted alternatives, including different thermodynamic phases. A rigid vessel fixes the total volume; it does not require each coexisting thermodynamic phase to retain its own volume.

For a constructed comparison at $300\mathrm{K}$, suppose one candidate has $1.5\mathrm{kJ}$ more internal energy but $6\mathrm{J}{\mathrm{K}}^{-1}$ more entropy than another at the same temperature and volume. Its difference isThus the candidate with higher internal energy has lower Helmholtz free energy. This is the practical reason for retaining the entropy term when selecting a state.

For equilibrium states within a smooth branch of fixed composition, the first law of thermodynamics gives $dU=TdS-pdV$, henceThese derivatives extract entropy and pressure; the differential is not an evolution equation for relaxation. If chemical species amounts $n_i$ vary, add ${\sum }_i{\mu }_{i}d{n}_i$, with molar chemical potentials ${\mu }_i$. A closed system can still change chemical species amounts through chemical reactions; the fixed-composition simplification then needs the conserved-inventory constraints of chemical equilibrium. See MIT 5.60, Lecture 14, fundamental equations and chemical potentials.

The minimum principle has a short explanation. At fixed volume, no boundary work occurs, so $Q=\Delta U$. A thermal reservoir at positive temperature $T$ loses entropy $Q/T$. Consequentlyby the Second law of thermodynamics. Here both endpoint states share the imposed temperature. See MIT 5.60, Lecture 12, criteria for spontaneous change.

The microscopic counterpart is $F=-k_{B}T\ln Z$ in the canonical ensemble, with Boltzmann constant $k_B$ and partition function $Z$. The partition function examples compute the balance between internal energy and entropy; the chemical equilibrium example already displays an ideal-mixture Helmholtz free energy minimum.

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.

An isolated system exchanges neither matter nor energy with its surroundings. Its total energy is fixed, while the Second law of thermodynamics requires its total entropy to be nondecreasing. Its internal energy alone is fixed when bulk kinetic energy and external potential energy also remain unchanged. For example, an initially spinning stirrer inside an insulated rigid vessel can lose kinetic energy while the contents gain internal energy, with no external input.

An isolated system need not yet be in thermodynamic equilibrium. Two enclosed bodies at different temperatures can exchange heat internally until they reach thermal equilibrium; their combined energy stays constant. Their internal transfer is absent from the outer boundary ledger. A vacuum-insulated container is only an approximation to isolation: residual thermal conduction and thermal radiation can still carry energy across its boundary.

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.

The laws organize the equilibrium and process constraints of thermodynamics: the zeroth law of thermodynamics makes thermal equilibrium transitive, the first law of thermodynamics expresses energy balance, the Second law of thermodynamics restricts entropy changes, and the third law of thermodynamics describes the zero-temperature limit. Their analogy with horizon mechanics becomes quantitative through Hawking temperature and Bekenstein-Hawking entropy.

A thought experiment about measurement, selective control and the Second law of thermodynamics. An active stochastic cooling system uses and dissipates resources while reducing beam deviations.

The Mach number is $\mathcal{M}=u/c_s$, comparing the flow speed relative to a disturbance with the local sound speed. For a body moving supersonically, spherical acoustic disturbances emitted earlier have a tangent envelope with Mach cone half-angle $\sin \theta =1/\mathcal{M}$. Signals cannot propagate arbitrarily far upstream. Compressive characteristics can converge and steepen into a shock wave, within which microscopic dissipation prevents a multivalued fluid state; supersonic motion alone does not require every flow to contain a shock.

Work in the shock frame with steady normal velocities $u_1,u_2>0$. Integrating conservation of mass, normal momentum and total energy over a thin pillbox, neglecting gravity across the thin layer, gives the Rankine-Hugoniot conditions for a perfect gas:The energy equation includes pressure work through specific enthalpy. The shock is adiabatic if negligible heat escapes the layer, but its irreversible dissipation means it is not isentropic.

Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and ${\mathcal{M}}_1=u_1/\sqrt{\gamma p_1/{\rho }_1}$. Momentum conservation gives $p_2=p_1+{\rho }_1{u}_1^2(1-1/\chi )$. Substitution in the energy equation produces the unshocked root $\chi =1$ and the compressive rootSince ${\rho }_1{u}_1^2/p_1=\gamma {\mathcal{M}}_1^2$, the strong-shock limit has negligible upstream thermal pressure compared with ram pressure. The Strong-shock Rankine-Hugoniot conditions giveFor unchanged composition, the ideal gas law implies $T_2/T_1=(p_2/p_1)/\chi$, soFor $\gamma =5/3$, the compression tends to four and $T_2\simeq 3\bar{m}{u}_1^2/(16k_B)$, where $\bar{m}$ is mean particle mass.

With constant specific heat $c_v$, entropy production in a perfect-gas shock is measured by the change in specific entropyThe strong-shock expression is thereforeThe exact jump is positive for every compressive shock with ${\mathcal{M}}_1>1$, in accordance with the Second law of thermodynamics; the asymptotic expression is only for ${\mathcal{M}}_1\gg 1$. There is a normalization issue in the printed description: this logarithmic formula with constant $c_v$ describes entropy per unit mass. The true entropy density is $\rho s$, so its difference is ${\rho }_2{s}_2-{\rho }_1{s}_1$, rather than the same logarithmic jump with a constant heat capacity per unit volume. A material parcel's irreversible entropy increase should be assessed using $\Delta s$.

Downstream gas is denser and hotter. For optically thin collisional emission the loss per unit volume is typically $n_e{n}_i\Lambda (T)$, whereas stored thermal energy scales approximately as $nT$. Compression can therefore reduce the radiative gas cooling time; excitation and ionization at the higher temperature can open additional cooling channels. The actual efficiency still depends on the astrophysical cooling function, composition and optical depth. In a radiative shock with separated scales, the temperature first rises across a narrow dissipative layer, then decreases through a much broader cooling region of characteristic length ${\ell }_{\mathrm{cool}}\sim u_2{t}_{\mathrm{cool}}$. Subsonic cooling usually brings further compression with approximately constant pressure. The immediate microscopic shock thickness and the macroscopic cooling length are distinct.

Let state $f$ be the fully cooled gas, with $T_f=T_1$ and the same composition. Set $a^2=p_1/{\rho }_1$, the isothermal sound speed squared. Across the entire shock plus cooling layer, mass and momentum are still conserved, but gas energy alone is not. With ${\chi }_f={\rho }_f/{\rho }_1$,The nontrivial isothermal shock therefore hasIf the upstream Mach number were defined with $a$ instead, this would be ${\rho }_f/{\rho }_1={\mathcal{M}}_{\mathrm{iso}}^2$. Retaining the distinction avoids losing the factor $\gamma$.

The gas's specific entropy rises irreversibly at the hot shock and then decreases as escaping radiation removes heat. Since the final temperature equals the initial temperature,This decrease for the gas is consistent with the Second law of thermodynamics: the emitted radiation and its eventual surroundings carry entropy away. It does not make the microscopic compressive shock reversible. The figure illustrates both the hot entropy jump and the subsequent cooling; its spatial cooling profile is schematic because no cooling law has been specified.

The entropy form of dry potential temperature follows from its parcel definition. The potential temperature is the temperature a parcel would reach if brought reversibly and adiabatically to a reference pressure $p_0$. For dry air of fixed composition, treated as an ideal gas with constant specific heat capacity,Per unit mass, the first law of thermodynamics, together with $\alpha =R_{d}T/p$ for specific volume, gives $\delta q=c_{p}dT-\alpha dp$. The reversible form of the Second law of thermodynamics then givesHence dry reversible adiabatic motion has $D\theta /Dt=0$. Pressure changes cause expansion cooling or compression warming, while the pressure factor in theta compensates exactly. Adiabaticity alone is insufficient if irreversible mixing or viscous dissipation produces entropy. Condensation or freezing supplies latent heating, so ordinary dry potential temperature need not be conserved in a moist adiabatically moving parcel.