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A geological dike is a sheet-like body or conduit formed when magma enters a fracture cutting across the surrounding rock. A vertical narrow dike can be modeled as a two-dimensional channel for coupled flow, heat transfer and wall melting.

The Drude model explains finite electrical conductivity by balancing charge carrier acceleration in an electric field against loss of directed momentum through collisions. Its useful surprise is that continuous microscopic motion can produce a constant macroscopic current density: the applied electric field continually restores the directed momentum lost to the material. The simplest version has one mobile species, constant number density $n$, electric charge $q$, mass $m>0$, and momentum relaxation time $\tau >0$, in a stationary, homogeneous, isotropic material. At zero magnetic field, its mean drift velocity and current density obeyFor electrons, $q=-e$ with positive elementary charge $e$, so ${\sigma }_0=n{e}^2\tau /m$. The electron drift velocity points against a constant electric field, whereas conventional current points along it. Random microscopic velocities can have zero mean without being zero individually; drift velocity is this population mean, rather than each particle's speed.^[ref]

The same first-order linear differential equation predicts both Drude current relaxation after switching an electric field and the Drude frequency response to a sinusoidal electric field. Together they let a measured electrical conductivity be interpreted in terms of a density, a mass, and a time scale; a single constant-field measurement determines only their combination $n{q}^2\tau /m$.

The original 1900 paper Zur Elektronentheorie der Metalle by Paul Drude sought a common account of electrical transport and heat transfer using mobile charged particles and ideas from the kinetic theory of gases. Its introduction, especially printed pp. 568–569, explains that comparing different transport phenomena could constrain the populations of free and bound charged particles. The modern momentum relaxation time formulation above is a useful presentation of the model, not a verbatim translation of the original notation or its historical assumptions about mass.^[ref]

An electrical calorimetry experiment asks whether a measured electrical work input accounts for a measured warming. The University of Iowa's demonstration uses an insulated flask containing water, an immersed heater, a thermometer, electrical meters and a timer. Its procedure compares the duration, electrical readings and temperature rise. The following low-power measurement design and synthetic readout develop that idea; they are not measurements reported by the demonstration. ^[ref]

Use a covered calorimeter with weighed water, an immersed resistor, a calibrated thermometer and gentle stirring. Put an ammeter in series with the heater branch and a high-input-electrical resistance voltmeter across the immersed resistor's terminals. Record $V(t)$, $I(t)$, temperature $T(t)$ and room temperature $T_a(t)$ before, during and after switching the heater on. Measure the electric potential difference at the heater, rather than at the supply: lead and contact electrical resistance outside the receiving boundary can consume power that never reaches the water. Account for voltmeter loading if the ammeter also includes the sensing branch's electric current.^[ref]

The electrical readout gives operating electrical resistance $R(t)=V(t)/I(t)$ when $I(t)\ne 0$, and power $P(t)=V(t)I(t)$. To test Ohm's law, obtain several pairs at the same temperature; one ratio alone cannot establish a linear response. For small electrical resistance, use four-terminal sensing to exclude lead drops and reverse electric current to test stable offsets from the thermoelectric effect. Use brief low-power resistance measurements if the intended result is the unheated value: the test itself causes Joule heating.

For a covered assembly at approximately constant pressure, with negligible evaporation, negligible endpoint mechanical energy changes and no phase transition, takeHere $c_w$ is the water's specific heat capacity at constant pressure, $C_{\mathrm{cal}}$ is the participating vessel, heater and sensor heat capacity, $Q_{\mathrm{loss},\mathrm{out}}$ is signed outward heat leakage, and $W_{\mathrm{stir},\mathrm{on}}$ is stirring work entering the assembly. The storage term approximates its enthalpy change; for a rigid fixed-volume receiver use the corresponding heat capacity at constant volume. Calibrate $C_{\mathrm{eff}}$ independently if the purpose is to test electrical input against thermal storage. Using the same heating run to define $C_{\mathrm{eff}}=E_{\mathrm{el}}/\Delta T$ and then claiming it verifies the equality would be circular.

For the synthetic trace, stipulate $V=5\mathrm{V}$, $I=1\mathrm{A}$, $R=5\Omega$, $C_{\mathrm{eff}}=500J/K$, room temperature $20{}^{\circ }\mathrm{C}$, and loss coefficient $G=0.5W/K$, with negligible stirring input. The heater runs for $120\mathrm{s}$, beginning at room temperature. The calorimeter heat-loss correction model isIt predicts $E_{\mathrm{el}}=600\mathrm{J}$, $\Delta T\approx 1.13080\mathrm{K}$, storage $565.398\mathrm{J}$ and outward loss $34.602\mathrm{J}$. Neglecting leakage undercounts deposited energy by about $5.77\%$; assuming no loss predicts $1.20\mathrm{K}$. After switching off, the excess temperature undergoes exponential decay with time constant $C_{\mathrm{eff}}/G=1000\mathrm{s}$. A cooling slope can estimate $G$ only with known $C_{\mathrm{eff}}$ and a credible single-temperature model.

Controls separate causes of a temperature rise. Check the electrical meters' zeros and thermometer calibration; keep water mass, immersion, lid and stirring fixed between runs. Record a heater-off stirred baseline to estimate sensor drift and stirring input. Repeat cooling runs over the same temperature range to estimate environmental heat transfer; a baseline taken only at room temperature cannot measure warming-dependent leakage. Vary electric current while keeping $R$ and initial thermal conditions comparable: corrected deposited energy should scale as $I^2$ for equal durations. An independent calorimeter calibration or known reference heat capacity makes this a quantitative test rather than a qualitative observation.

Finite sensor response, incomplete thermal equilibrium, changing $R(T)$ or heat capacity, evaporation, and imperfect loss corrections limit the inference. A final temperature alone cannot distinguish smaller electrical input from larger heat leakage or a larger participating heat capacity. Agreement tests the stated assembly and its energy accounting; it does not show that all electrical work in arbitrary devices becomes heat.

Historically, the Royal Society's archive description of James Prescott Joule's manuscript dated 13 October 1840 records a coil of wire in a jar of water, a thermometer and a galvanometer, and summarizes his inference that warming depends on electrical resistance and the square of electric current intensity. This electrical experiment precedes Joule's paddle-wheel experiment; the input here is electrical work, whereas the paddle supplies mechanical work. The catalogue, rather than the inaccessible manuscript images, supports this account. ^[ref]

An explicit lumped model makes the cancellation understandable. Treat the receiver as having uniform temperature $T$, heat capacity $C$, heater power $P_h$ deposited in the receiver, net parasitic input $P_{\mathrm{par}}$, and outward heat transfer $Q_{\mathrm{out}}(T,T_b)$ to a reference at temperature $T_b$. With incident radiant flux $P_{\mathrm{in}}$ and radiometric absorptance $a$, conservation of energy givesThe dark state has no admitted beam. Compare settled dark and light states, denoted $d$ and $l$; each has $dT/dt=0$. Subtracting their balances giveswhere $\Delta Q_{\mathrm{out}}=Q_{\mathrm{out},l}-Q_{\mathrm{out},d}$ and $\Delta P_{\mathrm{par}}=P_{\mathrm{par},l}-P_{\mathrm{par},d}$. At identical $T$ and $T_b$, with unchanged surroundings and parasitic heating, both correction differences vanish. Thus $P_{\mathrm{abs}}=\Delta P_h$, and $P_{\mathrm{in}}=\Delta P_h/a$. Notice the signs: extra stray thermal radiation admitted with the beam also reduces heater power, so it must be subtracted from that reduction before attributing the result to the intended beam.

Measure the heater's own terminal voltage $V_h$ and electric current $I_h$: for steady electrical heating, $P_h=V_h{I}_h=I_h^2{R}_h$, where $R_h$ is its electrical resistance. A supply voltage that includes wire voltage drops counts Joule heating in the leads too; only the portion deposited in the receiver belongs in this balance. Changes in that deposited lead heat belong in $\Delta P_{\mathrm{par}}$. Optical-electrical heating non-equivalence adds a further instrument-specific correction when the uniform-temperature model is insufficient.^[ref]

At fixed volumes and material inventories, two internally equilibrated bodies that exchange only heat within an isolated system have fixed total internal energy $E=E_1+E_2$. Their total entropy isUsing $1/T_i=(\partial S_i/\partial E_i{)}_{V_i,N_i}$, an interior stationary point obeysThus entropy stationarity requires matching thermodynamic temperatures. Positive values of heat capacity at constant volume make this a strict local maximum becauseFor bodies with positive heat capacities throughout the accessible range, this concavity gives at most one interior maximum. This is the stability check behind the equal-temperature condition; see entropy concavity and positive heat capacity. ^[ref]

For thermal equilibration of two finite bodies, entropy from heat capacity givesBody 1 gains about $0.22314355\mathrm{kJ}{\mathrm{K}}^{-1}$ of entropy, while body 2 loses about $0.19361556\mathrm{kJ}{\mathrm{K}}^{-1}$. Total energy is conserved even though total entropy increases. The entropy calculation uses reversible comparison paths between the endpoint states; actual exchange across a finite temperature difference is irreversible.

To construct the plot, vary $T_1$ while enforcing the energy constraint, rather than holding $T_2$ fixed. In numerical units of kelvin, $T_2/\mathrm{K}=(1500-T_1/\mathrm{K})/3$, and the plotted function isHere $\mathcal{S}$ means $S_{\mathrm{tot}}-S_{\mathrm{tot},\mathrm{initial}}$, so its zero is a reference choice. The chain rule yieldsThe unique maximum is at $T_1=T_2=375\mathrm{K}$. The physical trajectory starts at $T_1=300\mathrm{K}$ and moves toward that maximum; the rest of the curve compares other possible energy partitions, not later times in the trajectory.

The contact model also proves the direction of change directly:For $G>0$ and positive temperatures, equality holds exactly at thermal equilibrium. The appealing link is that a dynamical heat transfer law moves the system uphill on the constrained entropy curve, while conservation of energy keeps it on that curve. This predicts both why the endpoint is stable and why the reverse separation into hot and cold bodies cannot happen spontaneously in the isolated system.

A finite heat bath has a finite whole-body heat capacity $C_b$ and therefore changes temperature when it absorbs heat $Q_b$. For a rigid closed system with no work, fixed composition and no phase transition, the first law of thermodynamics givesIf the heat capacity at constant volume $C_b$ is constant throughout this temperature interval, the integral becomes $Q_b=C_b(T_{b,f}-T_{b,0})$. This statement concerns an internal energy difference; $C_b{T}_b$ need not be the body's absolute internal energy.

For an original example, put a rigid body 1 and a finite bath 2 into thermal contact inside an isolated system. Stipulate constant whole-body heat capacities $C_1=100\mathrm{J}{\mathrm{K}}^{-1}$ and $C_2=900\mathrm{J}{\mathrm{K}}^{-1}$, with initial temperatures $T_{1,0}=350\mathrm{K}$ and $T_{2,0}=290\mathrm{K}$. Each body remains internally uniform; the connection stores negligible energy; no external work, matter exchange or phase transition occurs. These are constructed model values. The argument in thermal equilibration of two finite bodies applies directly: equal final temperatures and conserved total internal energy giveThus the heat $Q$ transferred from body 1 to bath 2 isBody 1 loses $5400\mathrm{J}$ of internal energy and bath 2 gains exactly that amount. Body 1 cools by $54\mathrm{K}$, while bath 2 warms by $6\mathrm{K}$. Equal heat transfers produce unequal temperature changes because the heat capacities differ.

The heat bath approximation can now be assessed against a chosen allowed temperature change $\delta T>0$. Constant $C_b$ requires $|Q_b|/C_b\le \delta T$, rather than merely calling the bath large. For this example, an allowed change of $0.1\mathrm{K}$ would fail because the computed change is $6\mathrm{K}$. The exact bath shift and its ideal-reservoir limit areHold $C_1$ and both initial temperatures fixed in this limit. The transferred heat then approaches $C_1(T_{1,0}-T_{2,0})=6000\mathrm{J}$. The finite bath receives less heat because its warming reduces the available temperature difference. For heat capacities that vary, use the two integrals in the energy balance instead of this weighted average.

Conservation of energy also permits imagining the reverse transfer, so it cannot explain the direction of equilibration by itself. Entropy from heat capacity gives the additional test, using reversible comparison paths between equilibrium endpoints:All ratios inside the natural logarithms use absolute temperatures. The actual exchange across unequal temperatures is irreversible, even though reversible paths calculate the endpoint entropy differences.

There is a short general proof of positivity. Put $w_i=C_i/(C_1+C_2)$ for positive constant heat capacities and positive initial temperatures. The strict concavity of the natural logarithm giveswith equality only for equal initial temperatures. In this proof all temperatures inside the logarithms mean their numerical values in one common absolute unit. This is Jensen's inequality applied to the same weighted average that conservation of energy selected. The direction of heat transfer and the stability of its endpoint also follow from entropy maximization under thermal contact.

Why must a conducting piston move slowly? A simple model couples the first law of thermodynamics to Newton's law of cooling. Assume a fixed amount of ideal gas with uniform temperature $T(t)$ and constant whole-sample heat capacity at constant volume $C_V$, a heat bath maintained at $T_b$, negligible friction, and mechanically slow motion. Let $G>0$ be the effective thermal conductance, the reciprocal of the thermal resistance, in $\mathrm{W}{\mathrm{K}}^{-1}$. ThenPositive heat transfer must offset expansion work to keep $T$ near $T_b$. A finite $G$ and nonzero piston speed therefore require a temperature difference. This uniform-temperature model neglects spatial temperature gradients, sensor response, cylinder heat capacity and any change of $G$ with piston position; it is an illustrative calculation, not a fit to the isothermal gas syringe experiment.

For a prescribed exponential volume change $V=V_i{e}^{\lambda t}$, define $\tau =C_V/G$, $s=t/\tau$ and $ϵ=nR\lambda /G$. The dimensionless temperature $\theta =T/T_b$ obeys a linear ordinary differential equation with initial value $\theta (0)=1$:The plotted expansion uses stipulated $C_V=(5/2)nR$ and stops at $V_f=2V_i$, at $s_f=\ln 2/[(5/2)ϵ]$. During expansion, the larger $ϵ$ produces the larger temperature deficit; once the piston is held fixed, the deficit undergoes exponential decay as $e^{-(s-s_f)}$. Reducing $|ϵ|$ makes the path more nearly an isothermal process, while increasing its duration. Reversing the piston with $\lambda <0$ produces a temperature excess and outward heat transfer; the small-$|ϵ|$ regime is the intended near-isothermal approximation. Mechanical slowness alone does not ensure sufficient heat transfer.

Firn is compacted, recrystallized old snow that has not yet become dense glacier ice. Its porosity affects meltwater infiltration and heat transfer.

Stipulate a model receiver held at $T=5.4\mathrm{K}$ with a reference at $T_b=5.0\mathrm{K}$. Choose $G=5\mathrm{mW}{\mathrm{K}}^{-1}$, so outward thermal conduction is $G(T-T_b)=2\mathrm{mW}$. Ignore other heat transfer, take $P_{\mathrm{par}}=0$ in both states, and stipulate perfect optical/electrical thermal equivalence. These values specify an original example; they are not measurements or claimed performance of a NIST instrument.

With the beam blocked, supply $P_{h,d}=2\mathrm{mW}$. With light admitted, the required heater power is $P_{h,l}=1.005\mathrm{mW}$. For stipulated radiometric absorptance $a=0.995$ at the chosen single wavelength,The illuminated receiver still gets $1.005+0.995=2\mathrm{mW}$ of heating. The remaining $0.005\mathrm{mW}$ of incident radiant flux escapes without heating it. Calling the measured heater reduction “incident power” would underestimate the stipulated incident radiant flux by $0.5\%$.

Two variations expose different pitfalls. First, if admitting the beam also adds $0.010\mathrm{mW}$ of stray heating while the same $0.995\mathrm{mW}$ is absorbed from the intended beam, the heater falls to $0.995\mathrm{mW}$. The apparent reduction is now $1.005\mathrm{mW}$; subtracting $\Delta P_{\mathrm{par}}=0.010\mathrm{mW}$ recovers the beam's absorbed power. Second, keep the original heater readings but stipulate a light-state receiver offset $\delta T=+1\mathrm{mK}$ at fixed reference temperature. The link carries an extra $G\delta T=0.005\mathrm{mW}$; the inferred absorbed power becomes $1.000\mathrm{mW}$, and incident radiant flux becomes $1.000/0.995\simeq 1.005025\mathrm{mW}$. This second variation deliberately violates exact fixed-temperature substitution.

The heater can supply only nonnegative power in this model. Consequently an absorbed beam above the available dark heater power cannot be compensated at the chosen setpoint merely by reducing the heater. The operating point, link, or admitted radiant flux must change; zero heater demand is a range boundary, not evidence of greater radiant flux sensitivity.

Fourier's law predicts the local conductive heat flux from the local temperature gradient. In a continuum approximation with isotropic thermal conductivity $k>0$, its vector form and one-dimensional form areThe heat flux $q$ has units $\mathrm{W}{\mathrm{m}}^{-2}$, the temperature gradient has units $\mathrm{K}{\mathrm{m}}^{-1}$ and the thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. The rate of heat transfer $\dot{Q}$ has units $\mathrm{W}$. These units distinguish a local material response from the total power crossing a surface.

The minus sign makes heat flow toward lower temperature: for $k>0$, $\mathbf{q}\cdot \nabla T=-k|\nabla T{|}^2\le 0$. In the worked bar in thermal conduction, $dT/dx=-200\mathrm{K}{\mathrm{m}}^{-1}$ therefore gives $q=+400\mathrm{W}{\mathrm{m}}^{-2}$. Reversing the end temperatures reverses the heat flux. A zero temperature gradient gives zero conductive heat flux in this model.

Fourier's law is a constitutive relation expressing measured material behavior; it is not a consequence of conservation of energy alone. It applies locally and does not require a steady state. The straight-line bar solution additionally assumes constant thermal conductivity, constant cross-sectional area, no sources or side losses, and fixed end temperatures. Anisotropic materials require the directional response described by the thermal conductivity tensor. The scalar $k$ used in the examples expresses the stated isotropy assumption. ^[ref]

Historically, Joseph Fourier's Théorie analytique de la chaleur was published in 1822. In Chapter II, article 112, he writes the heat crossing a spherical surface as the product of thermal conductivity, elapsed time, surface area and the negative radial temperature gradient; article 113 balances the incoming and outgoing heat to obtain the heat equation for a sphere. The enduring idea is this connection between a local flow rule and a global prediction of changing temperature. ^{[ref] [ref]}

A high-Rayleigh number closure $\mathrm{Nu}=C{\mathrm{Ra}}^{1/3}$ gives $q=C{k}_T[g\beta /(\nu \kappa ){]}^{1/3}(\Delta T{)}^{4/3}$, independent of layer depth. It assumes boundary-layer-controlled heat transfer, a well-mixed interior, fixed material properties and a fixed Prandtl number regime. It is a model, not a universal asymptotic law for every turbulent convection regime or for an impulsive initial conductive transient.

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.

In the ideal kinetic friction model, the loss of combined mechanical energy of the two contacting bodies becomes thermal internal energy at rateThis follows by adding the contact powers $f_{k}v$ and $-f_{k}U$ from Newton's third law. It depends on contact slip, whereas the contact power received by either body separately depends on its velocity in the chosen inertial frame. Thus positive frictional heating is compatible with increasing kinetic energy of one body, as the moving-belt example in kinetic friction shows. Heating need not cause a temperature rise if thermal energy escapes by heat transfer.

William Francis Giauque's 1949 lecture, Some Consequences of Low Temperature Research in Chemical Thermodynamics, credits Peter Debye with an independent proposal. It describes gadolinium sulfate octahydrate in a jacket immersed in liquid helium, with an external solenoid. Exchange gas conducts away heat during magnetization; evacuation isolates the sample before field reduction. A liquid-nitrogen shield reduces radiative heat transfer. Its photograph identifies the 1933 apparatus operated with Duncan Peck MacDougall. An inductance bridge reads the sample's magnetic susceptibility; measuring coils remain outside the evacuated jacket in liquid helium. Increasing magnetic susceptibility indicates cooling, but converting it to temperature using Curie's law becomes approximate as that law fails.^[ref]

The measurement logic deserves a separate check. A changed inductance alone is not an absolute temperature calibration: the sample geometry, applied measuring magnetic field, and constitutive response enter the inference. A modern control would compare the susceptibility-based thermometer with an independently calibrated thermometer over their common range, check recovery after repeated sweeps, and measure subsequent warming to bound incoming heat. These are proposed controls derived from the readout principle, rather than additional procedures attributed to the 1933 experiment. The historical working salt has a richer magnetic level structure than the illustrative two-level paramagnetic cooling model.

A heat bath, or thermal reservoir, is an idealized system so large that exchanging a finite amount of heat does not appreciably change its temperature.

The ideal heat bath supplies a fixed absolute temperature $T_b$ while its internal energy changes through heat transfer. A real body's suitability as a heat bath depends on its heat capacity relative to the exchange being studied, and on how quickly its internal temperature gradients disappear. Large physical size alone supplies neither condition. A finite heat bath makes the approximation quantitative; a regulated calibration bath instead uses feedback control to maintain its temperature by exchanging energy with external equipment.

For a weakly coupled system that exchanges energy but no matter with an ideal heat bath, the canonical ensemble assigns the Boltzmann distributionHere $E_n$ is a microstate's energy, $k_B$ the Boltzmann constant and $Z$ the partition function. The fixed temperature removes the need to track every exchange with the environment. A substantially changing bath temperature no longer gives this one fixed Boltzmann distribution.^[ref]

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 current is a rate of heat transfer across a specified boundary. Its sign requires a chosen direction; unlike heat flux, it is not a rate per unit area.

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.

Heat-engine calorimetry measures heat transfers so that observed work can be compared with the correct energy input. Measuring motion alone cannot supply the denominator of thermal efficiency.

The institutional Laboratory 16: Stirling Engine describes a modern teaching arrangement: an ethanol lamp drives a Stirling engine coupled to an electrical output device, which supplies a resistor immersed in water in a calorimeter. The prescribed readouts include fuel mass consumed, the pressure-volume loop, electrical readings, and water temperature. The sequence separates fuel input, gas-cycle work, electrical output, and Joule heating of the water. Warming of the water is the final useful output; it is not the engine's rejected $Q_c$. The handout distinguishes warm-up from steady operation and asks students to compare the final warming with direct heating by the lamp. It gives a procedure, not a table of measured efficiencies. Its pp. 1–3 therefore establish an apparatus and measurement design without supplying a new experimental result.

For a proposed calibration, use a known input of electrical work to determine the effective heat capacity $C_{\mathrm{eff}}$ of the water, vessel and immersed parts; stir to make the measured temperature representative. A matched run with no input estimates exchange with the room. For constant $C_{\mathrm{eff}}$ and no phase transition, the resulting model iswhere $Q_{\mathrm{leak},\mathrm{out}}$ is the signed net heat lost to the room during the measurement. Here $E_{\mathrm{in}}$ is supplied electrical work when the resistor lies inside the calibration boundary; for a separate external heat input it is the received heat. These are proposed controls, not controls reported in that handout. Slow cooling, changing material properties and imperfectly matched room exchange limit this inference; the approach to thermal equilibrium and heat capacity matter as much as the thermometer reading.

To measure the actual engine-input $Q_h$, one must instrument its hot-side transfer separately, or close a calibrated energy balance around a specified heater boundary. A heater's electrical work includes changes in its stored internal energy and heat transfer that bypasses the engine. At periodic operation a cold-side flowing-water measurement can instead estimate ${\dot{Q}}_c=\dot{m}{c}_p(T_{\mathrm{out}}-T_{\mathrm{in}})$, if mass flow rate $\dot{m}$ and specific heat capacity at constant pressure $c_p$ are known, with negligible storage and other heat exchange in that measurement boundary. It measures only the rejection captured by that stream. Include all remaining transfers before using the residual to infer ${\dot{Q}}_h=P_{\mathrm{out}}+{\dot{Q}}_c$. This measurement discipline makes a comparison with Carnot's theorem meaningful.

Heat flux density is a vector $\mathbf{q}$ describing the rate of heat transfer per unit area, measured in $\mathrm{W}{\mathrm{m}}^{-2}$. Through a planar face of area $A$ with uniform normal component $q$, the signed rate is $\dot{Q}=Aq$ in $\mathrm{W}$; positive $q$ means heat travels in the chosen positive normal direction. Fourier's law determines the conductive contribution. In a bar with a constant cross-sectional area, steady thermal conduction without internal sources or side losses makes both $q$ and $\dot{Q}$ constant along the bar. Constant $\dot{Q}$ need not mean constant $q$ when the cross-sectional area varies.