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Calorimetry measures heat transfers. A calorimeter, a calibrated heat capacity and an accounting of unwanted heat transfer connect an observed temperature change to an energy balance, as in heat-engine calorimetry and heat-pump calorimetry.

The central idea is to measure a response that is easier to observe than the transfer itself: a known heat capacity turns a temperature change into an energy estimate. Mixing calorimetry can determine an unknown specific heat capacity, while a chemical reaction can be studied by measuring the response of its surroundings. The useful observable is a change in temperature, together with a stated thermodynamic system boundary; a hot object does not contain a separately conserved store of heat.

For a closed system with unchanged bulk kinetic energy and potential energy, let $Q_{\mathrm{in}}$ be net inward heat and $W_{\mathrm{on}}$ net inward work. The first law of thermodynamics givesIf the boundary includes a hot sample, receiving water and the participating vessel, their mutual heat transfer is internal. With no external input, the total storage change is minus the signed outward heat leakage, not the positive amount transferred from sample to water. A calorimeter calibration with an immersed resistor instead supplies electrical work across this boundary; Joule heating converts it into internal energy. Choosing only the receiving water as the boundary changes which transfers count as external.

The mechanical constraint decides which storage quantity the readout represents. For a closed system at fixed volume, with no other work, $Q_V=\Delta U$. At constant external pressure, with only expansion work $p\Delta V$, $Q_p=\Delta H$, where $H=U+pV$ is enthalpy. Thus the corresponding warming balances use heat capacity at constant volume or heat capacity at constant pressure. Electrical or shaft work must be included separately when present. These endpoint balances do not require the actual exchange to be a reversible thermodynamic process.

A covered cup open to atmospheric pressure approximately realizes the second constraint when escaping material is negligible. A sealed rigid reaction vessel realizes the first; its surrounding water, vessel and immersed parts form the calibrated receiver. The reaction's outward heat equals the receiver's gain only after losses and ignition input have been accounted for. The distinction concerns the reaction boundary, even if its receiver is open to atmospheric pressure.^[ref]

A historical quantitative result appears in James Prescott Joule's On the Mechanical Equivalent of Heat (1850): Table IX gives $772.692$ foot-pounds of mechanical work per warming of one pound of water by one degree Fahrenheit, with weighing reduced to vacuum. His final rounded statement specifies water between $55$ and $60$ degrees Fahrenheit. He acknowledged residual vibration and sound as possible upward bias. This is an original reported conversion result, not the definition of the modern joule; the apparatus and resting controls are described in Joule's paddle-wheel experiment.^[ref]

Heat is energy transferred across a thermodynamic system's boundary because of a temperature difference. It is measured in joules, like work and internal energy. A hot cup transfers heat to cooler surroundings; its temperature describes its state, while the transferred heat describes a process. The cup stores internal energy, rather than a substance called heat. ^[ref]

For a closed system with unchanged bulk kinetic energy and potential energy, choose $Q>0$ for heat entering and $W>0$ for work done by the system. The first law of thermodynamics givesHere $U$ is the stored internal energy. For example, a sealed piston cylinder receiving $20\mathrm{kJ}$ of heat and delivering $5\mathrm{kJ}$ of work gains $15\mathrm{kJ}$ of internal energy. A closed system retains its matter while exchanging energy; it need not be an isolated system. ^[ref]

Unlike internal energy, heat is not a state function: different paths between the same equilibrium states can transfer different $Q$, with compensating differences in work. The symbol $\delta Q$ records this dependence on the path; there is no stored quantity whose difference is always $Q$. The two paths in first law of thermodynamics illustrate this directly. Supplied heat can raise temperature as sensible heat, change a phase fraction through latent heat, or help supply expansion work. Measuring temperature alone therefore does not determine transferred heat.

For a reversible thermodynamic process, $\delta Q_{\mathrm{rev}}=TdS$, where $T$ is absolute temperature and $S$ is entropy. This identification requires reversibility: an irreversible adiabatic process has $Q=0$ and can nevertheless increase entropy. Likewise $Q=\Delta H$, with enthalpy $H$, requires the particular fixed-pressure conditions in enthalpy. These distinctions make heat useful as a precise boundary-transfer quantity in heat engines, refrigerators and heating calculations.

Historically, James Prescott Joule's paddle experiments showed that stirring water with input work could raise its temperature, helping establish conservation of energy. The lesson is still useful: a temperature rise can result from work even without incoming heat. ^[ref]

An extensive quantity measures how much of a thermodynamic system there is; an intensive quantity describes its state independently of that amount. The clean comparison is replication: take independent identical systems at the same thermodynamic equilibrium, with the same substance, composition and thermodynamic phase. Combining their accounting boundaries multiplies internal energy $U$, entropy $S$, volume $V$, mass $m$ and amount of substance $n$, while leaving temperature $T$, pressure $p$ and chemical potential $\mu$ unchanged.^[ref]

For $\lambda$ copies the ledger isAn integer $\lambda$ counts actual copies; a positive real $\lambda$ describes the corresponding bulk scaling law. Copies can remain separated: replication need not involve removing a partition, changing a container or performing any work. Distinguish $n$, the amount of substance measured in moles, from $N=n{N}_A$, the dimensionless count of specified entities, where $N_A$ is the Avogadro constant in ${\mathrm{mol}}^{-1}$. For example, $2\mathrm{mol}$ contains $1.204428152\times 10^{24}$ specified entities; tripling it gives $6\mathrm{mol}$ and $3.613284456\times 10^{24}$ entities. This uses the exact Avogadro constant $N_A=6.02214076\times 10^{23}{\mathrm{mol}}^{-1}$, which does not triple. In the differential $dU=TdS-pdV+\mu dn$, $\mu$ is a molar chemical potential in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$; the coefficient of $dN$ instead has units of energy per entity.

As a direct check, the ideal gas law $pV=nRT$ givesFor a monatomic ideal gas with only translational degrees of freedom active, internal energy of an ideal gas gives $U=3nRT/2$, so $U^{\prime }=\lambda U$ at unchanged temperature. Here $R$ is the molar gas constant. The scaling distinction lets a small sample supply an equation of state for a much larger amount: material state and inventory become separate questions.

Replication does not mean multiplying every linear dimension by $\lambda$: a geometrically similar container then has ${\lambda }^3$ times the volume. Nor does it mean combining samples at different states. Equal temperature establishes thermal equilibrium but does not imply equal pressure, composition or mass density. Density of a composite system makes this distinction concrete.

Internal energy $U$ is the energy stored in the microscopic degrees of freedom of a thermodynamic system: random molecular kinetic energy, internal molecular excitations, and interaction potential energy. Its value requires a specified system and an energy reference. In thermodynamic equilibrium, internal energy is a state function; it records the present state, without remembering whether a previous change came from stirring, compression or heat transfer. This makes internal energy the common accounting quantity connecting mechanical experiments to thermal ones.^[ref]

Heat and work describe energy crossing the system boundary during a process. They are not separately stored contents of the system. For a closed system whose bulk kinetic energy and external potential energy have unchanged endpoint values, the first law of thermodynamics givesHere $Q_{\mathrm{in}}$ is positive for incoming heat, and $W_{\mathrm{on}}=-W_{\mathrm{out}}$ is positive for net work done on the system. Two routes between the same equilibrium states have the same $\Delta U$, although their heat and work can differ. The infinitesimal notation $dU=\delta Q_{\mathrm{in}}+\delta W_{\mathrm{on}}$ distinguishes a state function change from path-dependent transfers.^[ref]

For a calculated example, take one mole of a classical monatomic gas in the ideal gas regime, at fixed volume, from $300$ to $310\mathrm{K}$. With stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, its internal energy change is $3nR\Delta T/2=124.71\mathrm{J}$. A purely thermal route supplies $Q_{\mathrm{in}}=124.71\mathrm{J}$ with $W_{\mathrm{on}}=0$. An ideal insulated stirring route supplies $W_{\mathrm{on}}=124.71\mathrm{J}$ with $Q_{\mathrm{in}}=0$, measured between equilibrium endpoints after the motion dies away. The same stored increment has two different transfer histories; neither history defines a separate kind of stored energy.

The familiar internal energy of an ideal gas is a useful special case. Neglect intermolecular interactions, hold the amount of substance $n$ and composition fixed, and suppose that $f$ independent quadratic terms per particle obey the classical equipartition theorem. Choosing the constant reference contribution as zero givesHere $T$ is absolute temperature, $R$ is the molar gas constant, and $C_V$ is the whole sample's heat capacity at constant volume. A monatomic gas has $f=3$ translational terms. A rigid diatomic molecule has $f=5$ when its two rotational terms are active and vibration is frozen; activating one harmonic oscillator vibration adds two terms, one kinetic energy and one potential energy, giving $f=7$. The counting is of quadratic contributions, so a vibrational coordinate is not just one contribution.^[ref]

This constant-$f$ formula applies only over a regime with those same active terms, classical translational statistics, nonrelativistic motion and unchanged composition. Quantum freezing, quantum degeneracy, chemical reactions or appreciable intermolecular interactions invalidate that simple counting. More generally, at fixed amount and composition an ideal gas can still have $U=U(T)$ with varying heat capacity at constant volume, so use $\Delta U={\int }_{T_1}^{T_2}{C}_V(T)dT$. A nonzero constant reference adds $U_{\ast }$ to $fnRT/2$ without altering $\Delta U$.

An isothermal compression reduces the volume of a thermodynamic system at fixed temperature. For a fixed-composition ideal gas compressed reversibly from $V_i$ to $V_f<V_i$, isothermal ideal-gas work givesThe surroundings supply positive work $W_{\mathrm{on}}$, and the gas rejects the same amount of heat to the heat bath. Compressing $1\mathrm{mol}$ at $300\mathrm{K}$ from $20$ to $10\mathrm{L}$ requires approximately $1.729\mathrm{kJ}$ of work input and rejects $1.729\mathrm{kJ}$ of heat. If that heat transfer cannot keep up, the temperature rises: the process departs from isothermal compression toward adiabatic compression. The finite-rate isothermal piston model explains the required small temperature excess.

An isothermal expansion increases the volume of a thermodynamic system at fixed temperature. For a reversible ideal gas path, the gas delivers positive work and absorbs positive heat, with $Q=W_{\mathrm{by},\mathrm{rev}}=nRT\ln (V_f/V_i)$ and $\Delta U=0$. The heat bath supplies the expansion energy; without sufficient heat transfer, doing work instead lowers the temperature. This transfer is the hot step of a Carnot cycle, and it is why an expanding gas need not cool when its surroundings continuously replenish its internal energy.

An insulated free expansion into an evacuated compartment is different. In the idealization with no heat exchange, no external work and no container energy change, $Q=W_{\mathrm{by}}=0$. Since internal energy of an ideal gas depends only on temperature, its final equilibrium temperature equals its initial value. The transient need not have a uniform temperature or pressure, so equal endpoint temperatures do not make it a reversible isotherm. Do not apply the logarithmic isothermal ideal-gas work formula to this irreversible process. Its entropy still increases by $\Delta S=nR\ln (V_f/V_i)$; entropy change in free expansion evaluates this state function difference using a hypothetical reversible path.^[ref]

For a real gas, even equal-temperature equilibrium endpoints need not have equal internal energy. An insulated free expansion can therefore change the temperature, while a controlled isothermal expansion must provide the heat needed for both its internal energy change and its delivered work. The internal energy at constant temperature article gives the Van der Waals equation example.^[ref]

An isothermal process changes a thermodynamic system while keeping its temperature constant. It need not keep its pressure, volume or internal energy constant, and it does not generally have zero heat transfer. An adiabatic process specifies zero heat transfer instead. For example, a slowly expanding ideal gas can absorb heat from a heat bath and deliver the same amount of work while its temperature stays fixed.

For a fixed amount of substance and composition, internal energy of an ideal gas depends only on temperature. With $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work delivered to the surroundings, the first law of thermodynamics therefore givesThis equality does not determine either transfer until the path is specified. Isothermal ideal-gas work gives the reversible value; an insulated free expansion has a different value even when its equilibrium endpoints have the same temperature. The closed system convention here excludes changes in bulk kinetic energy and external potential energy.

A conducting cylinder in a large, stirred heat bath approximates an isothermal process when the piston moves slowly enough for thermal conduction to keep up. A quasistatic process keeps the gas close to internal thermodynamic equilibrium; a reversible thermodynamic process additionally requires negligible friction and vanishing driving differences in pressure and temperature. Exact equality of the gas and heat bath temperatures gives no finite heat transfer rate through a finite thermal resistance. The reversible description is a limit of arbitrarily small differences and arbitrarily slow motion, illustrated by the finite-rate isothermal piston model.^[ref]

For a real gas, internal energy at constant temperature can change with volume because molecular interaction potential energy changes. In one smooth thermodynamic phase of a fixed-composition simple compressible system,Thus an isothermal process generally has $Q=\Delta U+W_{\mathrm{by}}$, with no reason to set $\Delta U=0$. The right-hand side vanishes for the ideal gas law, but not generally for another equation of state.^[ref]

Two independently available parts of a thermodynamic system are noninteracting in the canonical description when their state space is a Cartesian product and their energy is additive, $E(a,b)=E_A(a)+E_B(b)$. Both conditions are needed for the canonical partition function to factor:An additive expression on a constrained set of allowed state pairs need not give this factorization. Relabeling the states to make their allowed labels independent can instead introduce an interaction into the energy. For two two-state labels, an additive energy necessarily satisfies the rectangular identity $E_{11}+E_{22}=E_{12}+E_{21}$.

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 refrigerator removes heat from the region we want to keep cold and releases heat to warmer surroundings. Here we consider a work-driven cyclic device: its useful output is the cold-side extraction $Q_c$, and its input is net work $W$. A food compartment is a familiar example; the same cooling objective applies when removing heat from a room. The coefficient of performance measures extraction per unit work, rather than the fraction of input energy converted into output. ^[ref]

Choose the complete machine as the thermodynamic system. Let $Q_c>0$ be heat entering from the colder side, $Q_h>0$ be heat delivered to the hotter side, and $W>0$ be net work supplied, all over the same complete cycle. Assume these are its only energy transfers, no matter crosses the outer boundary, and there is no net change in stored energy. Returning to the initial state gives $\Delta U=0$, so the first law of thermodynamics readsThe heat leaving the warmer side includes both the extracted heat and the supplied work. For instance, stipulated transfers $Q_c=3\mathrm{kJ}$ and $W=1\mathrm{kJ}$ require $Q_h=4\mathrm{kJ}$; the cooling coefficient of performance is $3$. This is an illustrative budget, not measured performance. At steady operation the same balance holds for average transfer rates, ${\dot{Q}}_h={\dot{Q}}_c+P_{\mathrm{in}}$, where $P_{\mathrm{in}}$ is input power.

This budget explains a useful counterexample: operating a refrigerator with its door open inside an otherwise isolated room cannot cool the room as a whole. If both heat transfers occur within the room and the electrical input comes from outside, the room gains $Q_h-Q_c=W$ each cycle. Locally cold air does not imply net cooling. The Second law of thermodynamics permits cold-to-hot transfer because work is supplied; it forbids a cyclic device whose sole effect is that transfer. See the Clausius statement of the second law.

A thermodynamic system for which the reversible mechanical work is entirely pressure-volume work. At fixed composition its internal energy is specified by entropy and volume; additional work modes require additional state variables.

A state function assigns a value to each equilibrium thermodynamic state of a specified thermodynamic system. Its change depends only on the initial and final states. Internal energy $U$, entropy $S$ and enthalpy $H=U+pV$ are state functions; here $p$ is pressure and $V$ is volume. Transferred heat $Q$ and work $W$ belong to a thermodynamic process, rather than to a state. A system stores internal energy; it does not store separate inventories of “heat received” and “work received”.^[ref]

Use the explicit convention $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work done by the thermodynamic system on its surroundings. For a closed system with unchanged endpoint bulk kinetic energy and external potential energy, the first law of thermodynamics readsThe notation $dU$ denotes an exact differential of a state function; $\delta Q$ and $\delta W_{\mathrm{by}}$ denote incremental transfers. Along a reversible thermodynamic process of a simple compressible system,where $T$ is absolute temperature. These are rules for particular thermodynamic processes, not definitions of stored heat or stored work. In particular, a quasistatic process with friction can remain near thermodynamic equilibrium while producing entropy.^[ref]

For any state function $F$ and two equilibrium states $A,B$, an integral of its exact differential givesThe proof is telescoping: add the changes $F(x_{i+1})-F(x_i)$ along a subdivided path, and all intermediate values cancel. The smooth limit gives the integral. Conversely, if an integrable differential form has zero integral around every accessible closed path in a connected region, define $F(B)$ by integrating from a reference state. Two routes differ by a closed path, so the definition is unambiguous. This argument needs every accessible loop, rather than a local test that might miss a hole in the domain; see path independence.

The same cancellation does not hold separately for heat and work. A thermodynamic cycle has $\Delta U=0$ but can have nonzero net transfers, with $\oint \delta Q=\oint \delta W_{\mathrm{by}}$. Nor does the fact that entropy is a state function make $Q/T$ its general change: the formula $dS=\delta Q_{\mathrm{rev}}/T$ uses a reversible thermodynamic process and its varying temperature.

This distinction makes endpoint calculations powerful. We can compute an internal energy, entropy or enthalpy difference along a convenient reversible reference path even when the actual thermodynamic process is irreversible. Its actual heat and work must still be obtained from its own boundary interactions. The appealing cancellation is visible in two reversible ideal-gas paths between the same states: a different mechanical output requires a matching difference in thermal input, while stored internal energy arrives at the same value.

Two thermodynamic systems are in thermal contact when their boundary permits heat transfer. They need not already be in thermal equilibrium: contact between unequal temperatures is precisely how equilibration can begin. Contact specifies an allowed exchange; equilibrium specifies its balance.

A thermodynamic system is in thermodynamic equilibrium when its macroscopic state has no spontaneous tendency to change through any exchange allowed by its constraints. For an isolated system, stable thermodynamic equilibrium maximizes entropy over the accessible states at fixed total energy and conserved material inventories. A wall can forbid an exchange, so always specify what can cross or move. Thermal, mechanical and chemical conditions answer different questions about that same constrained state. ^[ref]

For example, two compartments separated by a rigid, impermeable wall that transmits heat can reach thermal equilibrium while retaining different pressures and compositions. Allowing an unloaded piston to move adds a mechanical equilibrium condition. Allowing one chemical species to cross adds chemical potential equality for particle exchange; allowing a chemical reaction adds its chemical equilibrium condition. A fixed barrier can therefore support a constrained thermodynamic equilibrium that ceases to be an equilibrium when the barrier is removed.

Uniformity is not the definition of thermodynamic equilibrium. An isothermal atmosphere in hydrostatic equilibrium can have a common temperature while its pressure and mass density vary with height: the pressure gradient balances gravity. In an external field, the exchange condition must include the relevant potential energy; for charged matter this leads to the electrochemical potential. Relativistic gravitational corrections to thermal equilibrium are described separately by the Tolman temperature law.

A steady state only says that the macroscopic description is constant in time. A bar maintained between hot and cold thermal reservoirs has steady heat flux and produces entropy, so it is not in thermodynamic equilibrium. Even matching temperatures at one perfect-contact interface can coexist with heat transfer through the surrounding material; matching those two local values does not establish equilibrium of the whole apparatus. See thermal conduction for that spatial transport problem.

The useful simplification is that equilibrium conditions predict an endpoint without computing every collision or flow during relaxation. The first law of thermodynamics supplies the conserved quantities, and the Second law of thermodynamics supplies the direction of change. For a system held at fixed temperature and volume, the corresponding criterion is minimization of Helmholtz free energy; at fixed temperature and pressure, it is minimization of Gibbs free energy, with the material and exchange constraints still imposed.

A state function used to describe a thermodynamic system with a chosen set of natural variables. Internal energy, enthalpy, Helmholtz free energy, and Gibbs free energy use different variables and serve different environmental constraints.

A thermodynamic process changes a thermodynamic system between states. A quasistatic process supplies equilibrium intermediate states and therefore a path on equilibrium coordinates. An irreversible thermodynamic process can have equilibrium endpoints while lacking a uniform intermediate temperature or pressure. Endpoint values still determine changes in a state function; computing heat and work requires the actual interaction history.

A thermodynamic system is the part of the physical world selected for an energy and matter balance. Draw its boundary before classifying an interaction as heat, work or transported matter. The boundary may move: a sealed piston encloses a closed system even while its volume changes. An isolated system additionally excludes all external energy transfers.^[ref]

Boundary choice changes the ledger. If an electrical heater warms water, choosing only the water counts incoming heat; including the heater counts electrical work entering through its leads. The combined internal energy records storage, without separate stored categories for the two transfer mechanisms.^[ref]

For a fixed-composition simple compressible thermodynamic system with smooth equations of state, write the internal energy as $U(T,V)$ and the enthalpy as $H=U+pV$. Differentiating $H$ at constant pressure and applying the multivariable chain rule givesSubtracting the heat capacity at constant volume proves the displayed identity. For an ideal gas, $(\partial U/\partial V{)}_T=0$ and $(\partial V/\partial T{)}_p=N{k}_B/p$, so $C_p-C_V=N{k}_B$. For a nonideal material the internal energy change with volume contributes as well as expansion work.

If two thermodynamic systems are each in thermal equilibrium with the same third system in the same equilibrium state, they are in thermal equilibrium with each other. Write $A\sim B$ for this relation. The zeroth law of thermodynamics supplies transitivity:Its symmetric relation property and self-equilibrium (reflexivity), together with transitivity, make thermal equilibrium an equivalence relation. Its equivalence classes can be labelled by a common empirical temperature; choosing an absolute numerical scale requires the further structure of thermodynamic temperature.^[ref]

The comparison assumes internally equilibrated macroscopic bodies, unchanged composition and mechanical constraints, and contact permitting only heat exchange. Use rigid, impermeable diathermal walls so that neither matter transfer nor boundary work spoils the test. The same state of $C$ matters: letting $C$ warm while measuring $A$, then cool while measuring $B$, does not supply the premise. Ordinary nonrelativistic laboratory thermodynamics is assumed; the Tolman temperature law treats gravitational qualifications. Equal temperatures establish the thermal condition, without requiring mechanical equilibrium or chemical equilibrium.

This explains why a thermometer can compare unlike materials: it supplies a reusable third system. A temperature is a property of an equilibrium state, while heat is energy transferred because of a temperature difference. Two bodies at the same temperature can have different internal energies and heat capacities. A thermometer therefore does not measure a body's stored amount of heat; the finite-body disturbance and waiting-time calculations in thermometer describe how making the comparison can change the state being measured.^[ref]

The name “zeroth” reflects logical order rather than the date of the first thermometer: this premise about thermal equilibrium precedes the already numbered first law of thermodynamics, Second law of thermodynamics and third law of thermodynamics. This is the naming explanation given in University Physics Volume 2 by Samuel J. Ling and others (§1.1, “Thermal Equilibrium”).^[ref]

The corresponding Zeroth law of black-hole mechanics states constancy of surface gravity over an equilibrium horizon; Hawking radiation identifies that quantity with physical temperature up to its universal conversion factor.