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Thermal diffusivity measures how rapidly thermal conduction smooths a temperature difference inside a material. Thermal conductivity measures the heat flux driven by a specified temperature gradient; thermal diffusivity also accounts for the heat capacity that must be warmed. A material can carry substantial heat yet change temperature slowly because it stores substantial internal energy per kelvin.

Write $\alpha =\kappa$ for thermal diffusivity, reserving $k$ for thermal conductivity. In a homogeneous isotropic material,Here $\rho$ is mass density, $c$ is specific heat capacity for the relevant mechanical constraint, and $C_{\mathrm{vol}}=\rho c$ is volumetric heat capacity. The familiar $k/(\rho c_p)$ uses specific heat capacity at constant pressure. For ordinary solids with negligible thermal expansion, the difference between the two constrained specific heat capacities is often neglected. A rigid fixed-volume model instead uses specific heat capacity at constant volume; a compressible moving fluid requires the accompanying work and transport terms. Thus the constraint is part of the approximation, not a choice made solely from the units. ^[ref]

Assume a stationary material in the continuum approximation, local thermal equilibration on times shorter than those resolved, no internal heating, no phase transition, and negligible convection and mechanical work. Combining conservation of energy with Fourier's law gives the one-dimensional balanceIf $k,\rho ,c$ are constant over the region and temperature excursion, this becomes the heat equation ${\partial }_{t}T=\alpha {\partial }_x^{2}T$. If $k$ varies, retain the differentiated product; replacing it with $k{\partial }_x^{2}T$ loses a term. Anisotropic materials need a thermal conductivity tensor. The scalar constant-$\alpha$ examples below deliberately exclude these complications. ^[ref]

For an original comparison, stipulate two model materials with equal thermal conductivity $k=2\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and equal mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, but specific heat capacities $c_A=1000$ and $c_B=4000\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. ThenAt the same temperature gradient, both have the same instantaneous heat flux by Fourier's law. At the same initial temperature profile and boundary conditions, material B takes four times as long to smooth a given spatial pattern. With fixed end temperatures, identical geometry and no sources, both eventually have the same steady profile and heat flux. The larger heat capacity changes the transient, not this steady state. These are stipulated parameters, not measurements of named substances.

Consider two rigid bodies forming an isolated system, with fixed composition and constant whole-body heat capacity at constant volume values $C_1=1\mathrm{kJ}{\mathrm{K}}^{-1}$ and $C_2=3\mathrm{kJ}{\mathrm{K}}^{-1}$. Their initial temperatures are $T_{1,0}=300\mathrm{K}$ and $T_{2,0}=400\mathrm{K}$. Assume each body remains internally uniform, their contact stores negligible energy, no phase transition occurs and no work is done or extracted. These are stipulated model values, not material measurements.

Because $d{U}_i=C_{i}d{T}_i$, write $U_i=a_i+C_i{T}_i$ with arbitrary fixed reference constants $a_i$. The conserved relative internal energy isThis is an energy reference convention, not a claim about actual heat capacities down to zero temperature. At thermal equilibrium the two temperatures equal $T_f$, so the first law of thermodynamics givesThe larger heat capacity pulls the common temperature closer to its initial value: body 1 warms by $75\mathrm{K}$, while body 2 cools by $25\mathrm{K}$.

To predict a timescale, stipulate a constant contact conductance $G=1/R_{\mathrm{th}}=0.1\mathrm{kJ}{\mathrm{s}}^{-1}{\mathrm{K}}^{-1}$, the reciprocal of the contact's thermal resistance. The rate of heat transfer into body 1 is $G(T_2-T_1)$, givingAdding proves conservation of $E$. Subtracting gives exponential decay of the temperature difference, with time constantand thereforeExact equality is approached as $t\to \infty$ in this ideal model. Changing $G$ changes the rate but not $T_f$: equilibrium and the speed of reaching it are separate predictions. Real bodies can have internal temperature gradients, changing heat capacities and contact losses, so the plotted $7.5\mathrm{s}$ is an assumed model timescale.

Two internally equilibrated systems are in thermal equilibrium if bringing them into contact that permits only heat exchange causes no net heat transfer. In ordinary nonrelativistic thermodynamics without appreciable gravitational temperature shifts, their thermodynamic temperatures are equal. Merely preventing heat transfer with insulation does not establish thermal equilibrium: two insulated bodies can have different temperatures. ^[ref]

A thermometer illustrates the definition. Its own temperature settles to the object's temperature after thermal contact, provided its effect on the object is negligible. The existing zeroth law of thermodynamics makes this comparison transitive: two objects separately in thermal equilibrium with the same thermometer are in thermal equilibrium with each other. This lets one calibrated scale compare different materials.

Equal temperatures do not imply equal internal energies or equal heat capacities. Nor does thermal equilibrium by itself imply mechanical equilibrium or chemical equilibrium; these require additional allowed exchanges. There is also a distinction between no net exchange in thermal equilibrium and a constant temperature maintained by equal incoming and outgoing heat. The latter can be a steady state with ongoing entropy production. Microscopic exchanges can continue at thermal equilibrium, but their opposing contributions balance on average.

Historically, OpenStax credits the name “zeroth law” to Ralph Fowler in the 1930s: the already numbered laws needed a logically earlier statement about thermal equilibrium. ^[ref]

For symmetric touching parabolic dispersions with constant density of states $D$ per energy band, including spin degeneracy, charge neutrality fixes $\mu =0$. The Fermi-Dirac distribution gives $N_c=D{k}_{B}T\ln 2$ and the thermal internal energy, including electrons and holes, is $\Delta U=({\pi }^2/6)D(k_{B}T{)}^2$. Thus $C=({\pi }^2/3)D{k}_B^{2}T$. Its large ratio to $N_c$ is compatible with the equipartition theorem because this excitation count grows with temperature and the heat capacity includes both species.

The derivative of internal energy at fixed deformation gradient and entropy, or of specific Helmholtz free energy at fixed temperature, defines this force. Its product with the internal-variable rate contributes mechanical dissipation to the Clausius-Duhem inequality.

A free energy is a thermodynamic potential that compares possible states under specified environmental constraints. For a closed system of fixed composition, treated as a simple compressible system with only pressure-volume work, stable thermodynamic equilibrium minimizes Helmholtz free energy at fixed temperature and volume, or Gibbs free energy at fixed temperature and pressure. The choice follows the experiment: a rigid sealed vessel and a sealed cylinder with a freely moving loaded piston impose different constraints. Both may exchange heat with a thermal reservoir. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19.

The useful insight is that internal energy alone cannot rank these alternatives: exchanging heat also changes the surroundings' entropy. Subtracting $TS$ packages that accounting into a property of the chosen system. Adding $pV$ additionally accounts for displacing the surroundings at fixed pressure. A free energy is a state function; supplied heat describes a transfer during a process. Under the stated constraints, supplied heat is $Q=\Delta U$ at fixed volume or $Q=\Delta H$ at fixed pressure, whereas neither $\Delta F$ nor $\Delta G$ generally equals $Q$.

A lower free energy identifies a thermodynamic preference, without specifying a timescale. A barrier can trap a system in a local minimum, producing metastability although a lower global minimum exists. The enzyme example illustrates the corresponding distinction for chemical reactions: changing the route can change the reaction rate while preserving the endpoint Gibbs free energy difference and chemical equilibrium.

For a simple compressible system at fixed composition and particle number, the internal pressure is the partial derivative of internal energy with respect to volume at fixed temperature. From $dU=TdS-pdV$ and the Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$, it has the measurable expression in the title. It is distinct from the mechanical pressure $p$.

For an ideal gas, $p=nRT/V$ gives ${\pi }_T=0$, so its internal energy depends only on temperature at fixed amount of gas. For the Dieterici equation $p=nRT\exp [-an/(VRT)]/(V-nb)$, direct differentiation givesThus an equation of state determines this volume sensitivity even without determining the full caloric function $U(T,V)$.

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.

Thermodynamics studies energy transfers and the constraints on macroscopic changes. Its vocabulary connects internal energy, entropy, temperature, pressure and volume to heat and work. The first law of thermodynamics balances energy; the Second law of thermodynamics decides which changes can occur spontaneously and limits how much work a heat engine can deliver. These principles let us bound performance before choosing a material or a detailed machine design. Applications include power generation, propulsion and refrigerators. ^[ref]

Start by choosing a system boundary. A closed system can exchange energy but no matter; an isolated system exchanges neither. A state function, such as internal energy or entropy, depends on the equilibrium state. A process specifies how that state changes. Thus a body's temperature is a property it has, whereas heat records energy crossing its boundary during a process. A body contains internal energy, rather than a stored quantity of heat.

Two insulated bodies initially at different positive temperatures approach thermal equilibrium when brought into thermal contact. Their combined energy is conserved, but their combined entropy increases. They do not spontaneously separate again into a hotter body and a colder one. This familiar asymmetry makes thermodynamics interesting: conserving energy leaves many imaginable changes available, and entropy supplies a further constraint. The same distinction underlies both entropy maximization under thermal contact and the thermal efficiency limit in Carnot's theorem.

For a film described by $dU=TdS+\gamma dA$, with material content fixed, its Helmholtz free energy obeys $dF=-SdT+\gamma dA$. Equality of mixed partial derivatives gives the Maxwell relation $(\partial S/\partial A{)}_T=-(\partial \gamma /\partial T{)}_A$. If $\gamma$ depends only on temperature, write ${\gamma }^{\prime }=d\gamma /dT$; thenHere $C_A$ is the heat capacity at constant area. For $C_A>0$ and ${\gamma }^{\prime }<0$, reversible adiabatic stretching cools the film, since $dT/dA=T{\gamma }^{\prime }/C_A<0$, although positive mechanical input can increase its total internal energy. If $\gamma ={\gamma }_0-\beta T$ with constant positive $\beta$ and constant $C_A$, the isentropic process satisfies $C_A\ln (T/T_{\ast })+\beta A=\mathrm{constant}$, where $T_{\ast }$ is an arbitrary fixed reference temperature. The work convention uses the total area conjugate to $\gamma$; a two-surface film described by projected area requires the corresponding factor of two consistently throughout.

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 the uniform noninteracting electron model with two spin states and kinetic energy of nonrelativistic particles $E=p^2/(2m_e)$, count all available one-particle states up to an energy $E\ge 0$:This extensive density of states has units of inverse energy; $g(E)/V$ is the density of states per unit volume. Availability and occupation are different: $g(E)$ is positive above $E_F$, but at zero temperature those states are empty. The occupied energy distribution is $g(E)f(E)$, with the Fermi-Dirac distribution reduced to $f(E)=1$ below $E_F$ and $f(E)=0$ above it. The convention at the exact boundary does not affect a continuum integral.

With the dimensionless quantity $x=E/E_F$, the occupied count per $dx$, divided by the total particle number, is $3\sqrt{x}/2$ for $0<x<1$. Its area is one. Weighting this by $E$ gives the internal energy and its mean per particle:Thus the highest occupied energy and the mean energy differ. For the Fermi energy of a model electron gas, $U/N\simeq 1.0155\mathrm{eV}$. At fixed $N$, $E_F\propto V^{-2/3}$, so the derivative $P=-(\partial U/\partial V{)}_N$ gives the electron degeneracy pressure $P=2n{E}_F/5$. The same state-counting geometry therefore predicts a nonzero pressure at zero temperature.

An ideal gas whose particles move in an area $A$ with two independent translational momentum components. In its dilute classical limit the one-particle translational partition function is $q=A/{\Lambda }^2$, where $\Lambda$ is the thermal de Broglie wavelength, and $Z_N=q^N/N!$. The translational internal energy is $N{k}_{\mathrm{B}}T$, and the surface pressure conjugate to area is $N{k}_{\mathrm{B}}T/A$. A mobile dilute adsorbed gas is an example; a fully occupied array of fixed adsorption sites has different state counting.

Use the independent spin-one-half two-level system with fixed $N$ distinguishable sites and Hamiltonian levels $E_{\pm }=\pm \mu B$, where $\mu >0$ is the magnitude of the field-axis magnetic moment and $B>0$ is a uniform externally controlled magnetic field. Assume positive temperature, no interactions between sites, and no appreciable lattice or load entropy. The two levels differ by $2\mu B$. With inverse temperature $\beta$, the canonical ensemble gives the one-site and total partition functionsHere the minus subscript denotes the lower energy $-\mu B$. Factorization of the partition function requires additive energy and independent sites; there is no particle-exchange factor $N!$ for these fixed distinguishable sites.

Differentiate the partition function, holding $B$ fixed when differentiating with respect to $\beta$. The internal energy, Helmholtz free energy, total field-axis magnetic moment $m$, and magnetic entropy areThe extensive moment $m$ is distinct from magnetization, the moment per unit volume. Direct substitution of $p_{\pm }$ into the Gibbs entropy gives the same $s(x)$ independently of the free-energy derivative.

At fixed $B$, the magnetic heat capacity isStrict decrease makes $s$ invertible on $x>0$. Constant magnetic entropy therefore fixes $x$, not $B$ or $T$ separately:The partition function may be used to label these equilibrium states even though the isolated cooling leg has no external heat bath. Actual thermalization of the lattice and load additionally requires a sufficiently fast spin-lattice relaxation; perfectly uncoupled spins by themselves cannot refrigerate an attached object.

The limiting cases are instructive. For $x\ll 1$, $s(x)=\ln 2-x^2/2+O(x^4)$, $m\simeq N{\mu }^{2}B/(k_{B}T)$, and $C_{B,\mathrm{mag}}\simeq N{k}_B{x}^2$. The moment response reproduces Curie's law. For $x\gg 1$, $s(x)\simeq (1+2x)e^{-2x}$ and the upper-state occupation is exponentially small. Taking $B\to 0$ at fixed $T>0$ gives $s\to \ln 2$; taking $T\to 0$ at fixed $B>0$ gives $s\to 0$. The origin has a path-dependent limit. Extrapolating a line of constant $x$ to $B=T=0$ does not construct a usable zero-temperature state or prove that a real material retains independent degenerate spins there.

Consider an original worked thermometer calibration, using the height $X$ of a liquid column responding through thermal expansion. Stipulate two equilibrated reference baths whose corrected thermometer readings are $0{}^{\circ }\mathrm{C}$ and $40{}^{\circ }\mathrm{C}$, in degree Celsius units. Suppose the corresponding column heights are $12\mathrm{mm}$ and $52\mathrm{mm}$. These are constructed inputs, not observed measurements. Assume the response is linear over this interval, with the same immersion and reading procedure throughout. Applying thermometer calibration givesFor $X=37\mathrm{mm}$, first subtract the zero-reference height, then divide by the full reference span: $(37-12)/(52-12)=25/40$. Multiplying by the $40{}^{\circ }\mathrm{C}$ span gives $t_C=25{}^{\circ }\mathrm{C}$, or $T=298.15\mathrm{K}$ in kelvin. Thus equal $37\mathrm{mm}$ readings in separately equilibrated samples test equal temperatures, regardless of how much internal energy those samples contain. The Celsius offset and modern absolute-scale definition are described in degree Celsius and kelvin.

The two endpoints determine the assumed line, but do not establish linearity. An independent middle reference and a return to the initial bath can test curvature and drift. If the height has an error of $0.2\mathrm{mm}$ while reference values and endpoints are held exact in this calculation, the inferred temperature error is $0.2{}^{\circ }\mathrm{C}$; that is only the height contribution, not a full measurement error budget. A boiling-water reference needs a measured atmospheric pressure correction, rather than an automatic $100{}^{\circ }\mathrm{C}$ assignment. The modern kelvin is not defined by either bath.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.

A hypothetical Van der Waals equation model illustrates positive and negative Joule-Thomson coefficient regions and a finite throttle that first warms, then cools. Use the molar convention already distinguished in Van der Waals equation:Here $v_{\mathrm{m}}$ is molar volume, $R$ the molar gas constant, $u_{\mathrm{m}}$ internal energy per mole, and $C_{V,\mathrm{m}}$ the stipulated constant molar heat capacity at fixed volume. The second equation is an additional caloric assumption: an equation of state alone does not specify the temperature dependence of internal energy. It satisfies the equilibrium identity $(\partial u_{\mathrm{m}}/\partial v_{\mathrm{m}}{)}_T=T(\partial p/\partial T{)}_{v_{\mathrm{m}}}-p=a_{\mathrm{m}}/v_{\mathrm{m}}^2$. The arbitrary energy reference is set to zero.

Define dimensionless variables $r=v_{\mathrm{m}}/b_{\mathrm{m}}$, $\theta =R{b}_{\mathrm{m}}T/a_{\mathrm{m}}$, $\pi =b_{\mathrm{m}}^{2}p/a_{\mathrm{m}}$ and $\eta =b_{\mathrm{m}}{h}_{\mathrm{m}}/a_{\mathrm{m}}$, where $h_{\mathrm{m}}$ is molar enthalpy. These differ from the critical-value scaling in Van der Waals law of corresponding states. The throttle calculation reduces toThus a prescribed inlet fixes $\eta$, and solving the first and last equations at the outlet pressure gives the final temperature. For an inlet $({\theta }_1,{\pi }_1)=(0.8,0.12)$ and outlet ${\pi }_2=0.025$, the model gives ${\theta }_2\simeq 0.752287$, a decrease of $0.047713$. The inlet $(2.2,0.12)$ instead gives ${\theta }_2\simeq 2.205786$, an increase of $0.005786$. The purple inlet $(1.45,0.30)$ initially warms but ends at ${\theta }_2\simeq 1.437585$ for the same outlet pressure. These are reduced toy values, not kelvin readings or experimental evidence. No physical gas parameters were fitted.

For a short inversion derivation, implicit differentiation of the Van der Waals equation at fixed pressure givesInsert this into $T(\partial v_{\mathrm{m}}/\partial T{)}_p=v_{\mathrm{m}}$ and substitute back into the equation of state. The Joule-Thomson inversion curve isPositive inversion pressure requires $r>3/2$. Its maximum pressure is $\pi =1/3$ at $r=3$, while the upper dilute limit is ${\theta }_{\mathrm{inv}}\to 2$. Using Van der Waals critical point values, the latter is $T_{\mathrm{inv},\max }/T_c=27/4$. The inversion locus depends on the equation of state; the chosen molar heat capacity additionally determines how much temperature changes along each finite throttle.^[ref]

The displayed region has $\theta >8/27$, above the model's critical temperature, and uses only $r>1$, $p>0$, $(\partial p/\partial v_{\mathrm{m}}{)}_T<0$ and positive $C_{p,\mathrm{m}}$. Subcritical calculations require attention to phase coexistence, metastability and the Maxwell construction; an unstable homogeneous continuation is not an equilibrium throttle endpoint. Even on a stable branch, this mean-field theory with constant $C_{V,\mathrm{m}}$ is a qualitative model, not an accurate inversion map for a particular real gas. Phase transitions, changing composition, appreciable heat leakage or bulk kinetic energy changes require a different property calculation or steady-flow energy balance. At phase coexistence, the smooth single-thermodynamic phase coefficient formula cannot replace a balance involving the phase fraction.

A vibrational partition function sums over molecular vibrational energy levels. A single quantum harmonic oscillator, with energies measured relative to its ground level and vibrational temperature ${\theta }_v=h\nu /k$, has $q_{\mathrm{vib}}=(1-e^{-{\theta }_v/T}{)}^{-1}$. Retaining its absolute zero-point energy multiplies this by $e^{-{\theta }_v/(2T)}$. That factor shifts the internal energy but contributes no entropy.

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.