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For a sinusoidal ground-temperature model whose coldest surface day is $t=0$, source temperature on that day is $T_0-T_A{e}^{-x}\cos x$, where $x=z/\delta$. The first and largest thermal maximum occurs at $x=3\pi /4$, not at infinite depth. This maximizes the winter coefficient of performance for a reversible pump with a fixed indoor temperature. With a heating load even about that day, the same optimum minimizes annually integrated reversible heating work. Excavation cost, finite heat extraction and real heat-pump behaviour can change the optimum.

For an isolated system, the second law requires $\Delta S\ge 0$. More generally, include all affected surroundings to obtainEquality describes the reversible limit; irreversible macroscopic processes produce entropy. A closed system can lose entropy by exporting heat, so nondecrease does not apply to that system alone. The Clausius statement of the second law forbids a cyclic device whose sole effect is transferring heat from cold to hot. The Kelvin-Planck statement of the second law forbids a cyclic device whose sole effect is converting heat from one thermal reservoir completely into work. "Sole effect" matters: a refrigerator moves heat from cold to hot by consuming work. ^[ref]

For direct transfer of $Q=1000\mathrm{J}$ from a thermal reservoir at $600\mathrm{K}$ to one at $300\mathrm{K}$, with no work extracted,The reverse transfer alone would have negative total entropy change. Both directions conserve energy, but only the hot-to-cold direction is spontaneous. The finite-body example under entropy replaces these constant temperatures by changing ones.

Friction, free expansion and heat transfer across a finite temperature difference all introduce irreversibility. Moving arbitrarily slowly is insufficient if friction remains. A reversible thermodynamic process must permit restoration of both the system and surroundings. This gives engineers a useful ideal benchmark: irreversibility spends an opportunity to obtain work, even though energy itself is conserved. ^[ref]

A secular tidal torque is the torque remaining after averaging the changing tidal interaction over fast orbital and rotational motion. The misaligned tidal deformation lets the companion's gravity change the spin angular momentum; the orbit receives the opposite torque. Tidal dissipation converts mechanical energy into heat, while total angular momentum is conserved in an isolated system. Thus the spin-orbit coupling exchanges angular momentum rather than destroying it. ^[ref].

For intuition near synchronous rotation on a circular orbit, hold mean motion $n$ and moment of inertia $I$ fixed and adopt the illustrative linear dissipative lawHere $\omega$ is spin angular velocity, $\theta$ is a marked surface direction in an inertial frame, and $K$ is a response coefficient, not a universal material constant. A faster spin is slowed and a slower prograde spin is accelerated. The solution and the associated mechanical power loss areThe orbital contribution to this power is $-n⟨\tau ⟩$, and the spin contribution is $\omega ⟨\tau ⟩$; their sum becomes heat. Holding $n$ fixed is a local approximation to the coupled evolution. Actual secular tidal torque depends on the deformation response and orbital eccentricity, so this simple law does not establish which spin-orbit resonance an eccentric body captures into.

Sensible heat is transferred heat associated with a temperature change within a single thermodynamic phase. By contrast, latent heat can change the phase fraction while the temperature stays fixed during equilibrium phase coexistence. Neither name denotes an additional stored substance: both describe how transferred heat participates in a process.

For fixed mass $m$ and composition, use the heat capacity appropriate to the heating constraint. Under the quasistatic process assumptions in heat capacity, with only pressure-volume work and unchanged bulk kinetic energy and potential energy, fixed-pressure heating givesHere $c_p$ is specific heat capacity at constant pressure; fixed-volume heating instead uses specific heat capacity at constant volume $c_V$. A cooling step has negative $Q_{\mathrm{sens}}$ with the sign convention in heat. A temperature difference has the same numerical value in units of the kelvin or degree Celsius. The integral follows by adding the small transfers $\delta Q=m{c}_p(T)dT$, rather than assuming a constant heat capacity.

As a stylized ice-to-water example for a fixed-composition closed system at fixed pressure $1\mathrm{atm}$, take $m=0.5\mathrm{kg}$, an initial temperature of $-10{}^{\circ }\mathrm{C}$, melting at $0{}^{\circ }\mathrm{C}$, and a final temperature of $20{}^{\circ }\mathrm{C}$. Assume constant specific heat capacity at constant pressure values $c_{\mathrm{ice}}=2.1\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $c_{\mathrm{water}}=4.18\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and latent heat of fusion $L_f=334\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. These are stipulated, rounded model values; ignore the container and heat losses, retain all the material, and assume only pressure-volume work under the quasistatic process conditions above. The three steps areOnly the first and third terms are sensible heat; the middle term is latent heat. The plateau begins after $10.5\mathrm{kJ}$ and ends after $177.5\mathrm{kJ}$ of total supplied heat. At $94\mathrm{kJ}$ total input, half the mass is liquid water and the temperature remains $0{}^{\circ }\mathrm{C}$. The equilibrium phase transition temperature depends on pressure; this example stipulates a single fixed pressure. ^[ref]

This split makes heating budgets easy to interpret: melting consumes $167/219.3\approx 76.2\%$ of the supplied heat in this model despite causing no temperature rise. Within either single thermodynamic phase, the heating curve slope is $dT/dQ=1/(m{c}_p)$; a larger heat capacity makes the slope shallower. A flat interval represents changing phase fraction, so it cannot be modeled by applying $Q=m{c}_p\Delta T$ across the phase transition. The complementary constructed example in latent heat tracks that phase fraction explicitly.

heat generation through muscle activity associated with shivering.

Specific heat capacity is heat capacity per unit mass:Its units are $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and the constraint remains part of the definition. In particular, specific heat capacity at constant volume and specific heat capacity at constant pressure are different responses. For homogeneous material, doubling the mass doubles its heat capacity while leaving its specific heat capacity unchanged. ^[ref]

As an illustrative material model, give a fixed-volume body mass $m=2\mathrm{kg}$ and constant specific heat capacity at constant volume $c_V=400\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. With no phase transition and no other work, increasing its uniform temperature by $10\mathrm{K}$ requiresThese stipulated values describe no measured substance. The right panel of the diagram in heat capacity shows the straight line $Q=m{c}_V\Delta T$: a temperature increase of $5\mathrm{K}$ takes $4000\mathrm{J}$ of heat. If the specific heat capacity at constant volume depends on temperature, use $Q=m{\int }_{T_1}^{T_2}{c}_V(T)dT$ instead. The corresponding fixed-pressure calculation uses specific heat capacity at constant pressure and gives $Q=\Delta H$ under the assumptions in heat capacity.

Choosing between specific heat capacity, molar heat capacity and volumetric heat capacity is choosing whether to compare equal mass, equal amount of substance or equal volume. A small sample with large specific heat capacity can still have less total heat capacity than a much larger sample.

Specific internal energy $e=U/M$ is internal energy per unit mass, measured in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. For a perfect gas with constant specific-heat ratio $\gamma$, choosing the constant reference as zero gives $e=c_{v}T=p/[(\gamma -1)\rho ]$, where $c_v$ is the specific heat capacity at constant volume, $T$ the temperature, $p$ the pressure and $\rho$ the mass density.

For a single-component fluid, the local first law of thermodynamics can be writtenThe material derivative follows velocity $\mathbf{u}$; $\mathbf{\tau }$ is the viscous stress tensor, $\mathbf{q}$ the heat flux density, and $r$ an external heating rate per unit volume. Compression supplies pressure work, the stress contraction accounts for viscous dissipation, and convergence of heat flux raises internal energy. Writing only $-p\nabla \cdot \mathbf{u}+Q$ assumes that viscous heating is negligible or included in $Q$. An insulated sheared fluid can warm even when no heat crosses its boundary.^[ref]

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.

For steady flow through a fixed control volume with one inlet and one outlet, the first law of thermodynamics and conservation of mass giveHere $\dot{m}$ is the mass flow rate, $\dot{Q}$ incoming heat per unit time, ${\dot{W}}_s$ outgoing non-flow work per unit time, $w$ the flow speed, $z$ elevation and $g$ gravitational acceleration. The specific enthalpy $h=u+p/\rho$ includes both the transported internal energy per unit mass $u$ and the pressure work needed to push volume through the boundary; $\rho$ is mass density. The inlet supplies that pressure work, while the outlet consumes it to displace the downstream fluid.^[ref]

For an insulated stationary throttle, require negligible heat transfer, no shaft or other non-flow work, and negligible changes in bulk kinetic energy and gravitational potential energy. Then $h_2=h_1$. With unchanged composition this is also equality of molar enthalpy. The upstream and downstream states must admit equilibrium pressure, temperature and property values, although the constriction itself is irreversible and can be far from thermodynamic equilibrium. Start-up storage of energy in the walls and fluid, heat leakage, or a large outlet speed invalidates the reduced balance. A nozzle can convert specific enthalpy into bulk kinetic energy even when it is insulated.

The same accounting for a fixed transported amount is $U_2-U_1=p_1{V}_1-p_2{V}_2$, hence $U_1+p_1{V}_1=U_2+p_2{V}_2$. This explains why dropping the two pressure work terms would give the wrong conserved quantity.^[ref]

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

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

A steam engine is a heat engine that converts part of the energy carried by steam into mechanical work. In a reciprocating design, expanding steam pushes a piston; valves admit and exhaust the working fluid so that the motion can repeat. A crank can convert the reciprocation into rotation. A steam turbine instead extracts work from a continuously flowing stream.

The boiler supplies heat to water, but the engine's useful output depends on pressure changes during expansion and exhaust. For an idealized quasi-static piston, boundary work is $W=\int pdV$. The net work over a cycle is the signed enclosed area on a pressure-volume diagram. Some input heat must be rejected: turning boiler heating into motion does not circumvent the Second law of thermodynamics.

Steelmaking produces steel by controlling the carbon content, other alloying elements and unwanted constituents of iron. A blast furnace first produces carbon-rich pig iron; a basic oxygen furnace then refines the molten material by oxidation and composition adjustment. An electric arc furnace instead supplies heat electrically to melt a charge that can include recycled steel, direct reduced iron and pig iron.^[ref]

The distinction between iron extraction and steelmaking matters: converting an iron oxide to metallic iron and adjusting an existing iron-based alloy to the required composition are different tasks. The Bessemer process is an important historical example of refining pig iron by an air blast; modern basic oxygen steelmaking uses an oxygen supply rather than ordinary air.

In stellar structure, stellar thermal equilibrium means that internal heating balances the surface luminosity and any energy loss through neutrinos, with negligible secular storage of thermal energy on the timescale considered. A star can satisfy this balance while its interior is hotter than its surface and heat flows outward. It is therefore different from equal-temperature thermal equilibrium between systems. Neglecting neutrino losses, the balance is $L_{\mathrm{surface}}=L_{\mathrm{heat}}$. In a slowly mass-losing donor star, a stellar thermal equilibrium radius sequence applies only when thermal adjustment is fast compared with the mass-loss timescale; rapid loss instead requires the adiabatic stellar radius response exponent.

A Stirling engine repeatedly heats and cools a confined gas, whose expansion and compression drive a moving mechanical load. This makes it a particularly accessible example of a heat engine: both the repeating motion and the two regions exchanging heat can be observed.

In the published Demonstration experiments with a Stirling engine by Christopher G Deacon and others, an electrical heater warms one end of a cylinder and flowing water cools the other at $12{}^{\circ }\mathrm{C}$. A moving displacer shuttles gas through copper wool that temporarily stores and returns internal energy. A laser spot traces a closed pressure-volume path on a screen: mechanical motion controls one coordinate, pressure the other. Calibrating both scales converts the enclosed area to work per thermodynamic cycle; multiplying by rotation frequency gives gas-cycle power. At $97\mathrm{W}$ heater input, the paper reports $4.4\mathrm{W}$ gas-cycle power, while a separate Prony brake measures $3.4\mathrm{W}$ shaft output. The cold-side temperature is controlled and two higher heater settings are compared. The gas has a nonuniform temperature, and heater energy also escapes to the room: neither output/input ratio measures $W/Q_h$ for an ideal two-thermal reservoir engine. These observations demonstrate work production and losses, not a direct measurement of the bound in Carnot's theorem. The inspected apparatus, readouts and results are on pp. 180–182, Figures 1–3 and Table 1.

Stochastic thermodynamics assigns work, heat and entropy production to individual fluctuating sample trajectories. It is useful for a small system, such as a trapped colloidal particle, whose exchanged energy is comparable to the thermal scale $k_{B}T$. The surroundings remain a thermal reservoir at temperature $T$, while the observed coordinates follow a stochastic process.^[ref]

For a one-dimensional optical trap with potential energy $U(x,\lambda )=k(x-\lambda {)}^2/2$, moving its centre $\lambda$ does incremental work $\delta W={\partial }_{\lambda }Ud\lambda$ on the particle. The coordinate change contributes ${\partial }_{x}U\circ dx=-\delta Q_{\mathrm{bath}}$, where the circle denotes the Stratonovich integral convention. The ordinary chain rule then gives the trajectory first law of thermodynamics, $dU=\delta W-\delta Q_{\mathrm{bath}}$. A particle ahead of the moving trap centre can give a negative instantaneous work increment, even during a protocol with positive mean dissipated work.

In the optical-trap experiment of G. M. Wang and collaborators, a recorded colloidal particle trajectory allowed short-time entropy-consuming realizations to be compared with entropy-producing ones. Repeating the same driving protocol measures a probability distribution rather than only a mean response.^[ref] The integral fluctuation theorem recovers nonnegative mean entropy production through Jensen's inequality, while allowing individual negative values.

A sun valve automatically restricts a gas light's supply in daylight and allows it again as illumination falls. In Gustaf Dalén's design, a blackened rod and three polished rods respond differently to sunlight. Their relative thermal expansion operates the gas valve through a membrane; a pilot flame permits relighting. The museum dates this design to 1907.^[ref]

The useful comparison is between two thermal responses. Both kinds of rod expand when warmed; a dark surface can absorb more incident radiant power, so its temperature rises more. Comparing their extensions can reject a common ambient-temperature change while responding to illumination. The mechanism converts light into heat, heat into a difference of length, and that difference into a change of gas flow. Differential thermal model of a sun valve makes those assumptions quantitative.

Temperature labels thermal equilibrium: two internally equilibrated bodies have the same temperature when contact permitting only heat exchange produces no net heat transfer. The zeroth law of thermodynamics makes the comparison transitive. If bodies A and B separately reach thermal equilibrium with the same thermometer without appreciably disturbing either body, they agree with each other. This is what lets a gas thermometer compare a metal block with water, despite their different internal energies and heat capacities. Insulation alone makes no such comparison: it can prevent heat transfer between bodies at different temperatures. ^[ref]

The zeroth law of thermodynamics establishes equality comparisons, but does not choose a numerical scale. Thermodynamic temperature supplies an absolute scale consistent with Carnot's theorem and entropy. For a simple compressible system at thermodynamic equilibrium with fixed particle number $N$ and composition,Here $U$ is internal energy, $S$ is entropy, and $V$ is volume; the second equality assumes the derivative can be inverted. Temperature is a thermodynamic conjugate variable to entropy and belongs to the intensive and extensive thermodynamic quantities as an intensive quantity. Its partial derivative tells us how much internal energy must change per unit change of entropy under the stated constraints. More general systems require their other mechanical and material variables to be held fixed too. ^[ref]

For an infinitesimal quantity $\delta Q>0$ of heat transferred from A to B, with no work or matter exchange, the equilibrium entropy changes giveFor ordinary positive temperatures, spontaneous heat transfer from $T_A>T_B$ increases total entropy; thermal equilibrium has $T_A=T_B$. This comparison uses changes in entropy, rather than the total internal energy contained in either body. A small hot object can have less internal energy than a large cool one. See entropy maximization under thermal contact for the equilibrium criterion and negative temperature for the special case of bounded energy spectra.

Express absolute thermodynamic temperature $T$ in kelvin, and a Celsius reading $t_C$ in degree Celsius:The offset is exact: $300\mathrm{K}$ corresponds to $26.85{}^{\circ }\mathrm{C}$, and $400\mathrm{K}$ to $126.85{}^{\circ }\mathrm{C}$. Their absolute temperature ratio is $4/3$; dividing their Celsius readings gives a different number. Thus gas thermometer pressure ratios and Carnot's theorem require thermodynamic temperature. The modern SI definition of the kelvin fixes the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$; a reference bath realizes or transfers the scale rather than defining that constant.^[ref]

Temperature is not universally the mean kinetic energy of a particle. In a classical, nonrelativistic ideal gas at thermal equilibrium, the equipartition theorem gives the mean translational kinetic energy relative to the gas's bulk motion:Here $m$ is particle mass, $\mathbf{v}$ its velocity, and $\mathbf{u}$ the mean flow velocity. In this regime, different particle masses at the same temperature have equal mean translational kinetic energies but different characteristic speeds. A moving gas can also carry bulk kinetic energy without being hotter in its rest frame. Internal molecular modes and quantum statistics require further care; the equipartition theorem has classical assumptions, whereas the equilibrium entropy definition applies more widely. ^[ref]

Temperature measurement infers temperature from a reproducible physical response of a thermometer. Examples include a liquid column's thermal expansion, a sensor's electrical resistance and the thermal radiation emitted by a surface. Calibration connects the response to specified reference temperatures or a physical temperature scale.

A contact probe must exchange heat with the object, so its reading depends on time and thermal coupling. A small probe can respond rapidly, while a large probe can noticeably disturb a small sample. Heat conducted along the probe stem and electrical self-heating can shift the reading away from the object's undisturbed temperature.

A useful immersion experiment is to insert a probe progressively farther into a well-stirred bath and wait after each movement. A reading that changes with immersion depth reveals a heat-leak or equilibration problem. Repeating a calibration does not by itself remove such measurement-condition errors.

The gas-only condition $\Delta U=0$ requires negligible heat exchange with the apparatus. If the insulated isolated system instead includes the walls, sensors and water, the first law of thermodynamics fixes their combined energy. Once everything returns to a common final temperature, a constant apparatus heat capacity $C_a$ givesDefine $A=U_g(T_1,V_2)-U_g(T_1,V_1)$, the gas's internal energy increase for the same expansion evaluated at the initial temperature. For a small temperature change, linearizing at the final volume givesHere $C_{V,g}$ is the whole gas sample's heat capacity at constant volume at the reference state. Compared with an isolated gas sample's $-A/C_{V,g}$, the common final signal is reduced by $C_{V,g}/(C_{V,g}+C_a)$. This energy accounting explains why a massive bath can conceal a real-gas effect. Fitzpatrick gives the corresponding apparatus correction for the Van der Waals equation. ^[ref]

For a toy calculation, stipulate $A=5\mathrm{J}$, $C_{V,g}=10\mathrm{J}{\mathrm{K}}^{-1}$ and $C_a=990\mathrm{J}{\mathrm{K}}^{-1}$, all constant in this example. The predicted cooling magnitude changes from $0.5\mathrm{K}$ for the gas alone to $0.005\mathrm{K}$ for the combined apparatus. An illustrative effective detection threshold of $0.01\mathrm{K}$ would miss the latter. These are chosen numbers, not Joule's measurements. In this approximation a null observation only bounds $|A|\lesssim (C_{V,g}+C_a){ϵ}_T$, where ${ϵ}_T$ is the detectable temperature change after allowing for drift and other experimental uncertainties; a display's smallest digit is not automatically this threshold.

The historical limitation was recognized explicitly: Joule and Thomson later estimated that the common apparatus in Joule's experiment would cool by only about $0.003\mathrm{K}$, explaining why the earlier external thermal effect was undetectable. This was a later calculation about the apparatus, not a measured cooling trace. ^[ref]

A temperature sensor response time specifies how long a temperature sensor takes to complete a stated fraction of its response to a change in the surrounding temperature. The fraction and the test conditions are part of the specification: a 90% response time and a 63.2% response time describe different points on the same record. Rapid immersion in a different-temperature liquid approximates the required step input.^[ref]

For a lumped thermal model with sensor heat capacity $C$, constant thermal conductance $G$ to a bath at $T_b$, and negligible internally generated heat, the energy conservation equation and its solution areThus the time constant $\tau$ is the time to complete $1-e^{-1}\approx 63.2\%$ of the change. The time to complete a fraction $f$ is $t_f=-\tau \ln (1-f)$, so $t_{90}=\tau \ln 10$, whereas the elapsed time between 10% and 90% is $\tau \ln 9$. For example, $\tau =3\mathrm{s}$ gives $t_{90}\approx 6.91\mathrm{s}$ and $t_{99}\approx 13.82\mathrm{s}$.

A two-bath step test of a temperature sensor tests this exponential function model rather than assuming it. Several internal heat capacity elements, a protective sheath, or electronic filtering can produce several response stages. The thermometry guide reports fast and slow components in tested platinum thermometer assemblies; a single quoted time constant need not describe their entire approach to equilibrium.^[ref]