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Assume a stationary temperature profile, spatially uniform mass density and radioactive heat generation rate, constant thermal conductivity, isotropic solid thermal conduction, and blackbody emission into effectively cold surroundings. The volumetric source isTotal radiogenic power equals the surface blackbody radiation loss. The Stefan–Boltzmann law givesFor steady radiogenic heating of a conducting sphere, the interior heat equation isRegularity at the centre removes the $1/r^2$ integration term and yields $dT/dr=-q_{v}r/(3\kappa )$. Matching $T(R)=T_S$ givesThe high central temperature is compatible with a cold surface because finite thermal conductivity requires a large temperature gradient to carry the accumulated interior heat. A nonzero background radiation temperature $T_{\mathrm{bg}}$ replaces $T_S^4$ by $T_S^4-T_{\mathrm{bg}}^4$ in the surface balance. The calculation neglects convection, stored transient heat, changing radioactive abundances, phase changes and any substantial external irradiation.

Choose $t_1=1/(\kappa k^2)$ to make the sketch explicit. The three profiles areThey have the same wavelength, zero crossings and phase, with amplitudes $A$, $0.3679A$ and $0.1353A$. Diffusion reduces the temperature gradient and amplitude; it does not advect or change this individual mode’s wavelength.

An upper atmospheric region above the tropopause in which shortwave heating can produce an atmospheric thermal inversion. Ozone on Earth, and methane and aerosols on giant planets, are examples of absorbers that help heat it. Its positive altitude temperature gradient makes it stable against homogeneous-gas thermal convection.

Combining geostrophic balance, $fu=-{\Phi }_y$, $fv={\Phi }_x$, with hydrostatic balance, ${\Phi }_p=-RT/p$, and commuting the horizontal and vertical derivatives proves the displayed equations. Horizontal derivatives hold pressure fixed. This relates vertical shear to horizontal temperature gradients away from the equator; the derivation requires nonzero Coriolis parameter and the validity of geostrophic balance.

A motivating real procedure is NIST's ambient thermometer verification. Its apparatus is a deep beaker of water, a calibrated reference thermometer, a test thermometer, and a magnetic stirrer with any heater switched off. Let the water approach room temperature, stir gently, immerse both sensing tips well below the surface, and wait before reading. NIST specifies about $20\mathrm{cm}$ of water, $10$–$15\mathrm{cm}$ immersion where possible, and an initial five-minute wait. Record test then reference, repeat in reverse order, and apply the reference's certificate correction. Subtract corrected reference from test: the difference estimates the test instrument's error at this bath temperature. Agreement between the two reading orders checks repeatability. ^[ref]

For example, suppose a test reads $24.7{}^{\circ }\mathrm{C}$, a reference reads $24.4{}^{\circ }\mathrm{C}$, and its certificate says to add $0.1{}^{\circ }\mathrm{C}$. The corrected reference is $24.5{}^{\circ }\mathrm{C}$, so the test error is $+0.2{}^{\circ }\mathrm{C}$: it reads high. These are illustrative readings. This verification tests a previously calibrated instrument at one point; it does not establish the two-point response used in two-point thermometer calibration example.

Additional controls follow from the thermal equilibrium premise. Keep tips away from walls and bottom, interchange probe positions to test a temperature gradient, and check that longer waits and modest immersion changes leave agreement unchanged within the required tolerance. A flat trace alone can be a steady state maintained by competing heat transfer paths. Five minutes is a procedural starting point, not a universal equilibration time. Excessive stirring can supply work and warm the bath, while probe stems can exchange heat with the room. These controls test whether both probes sample the same state; neither one-point agreement nor this experiment proves the zeroth law of thermodynamics for every material and condition.

A regulated bath holds its temperature near a chosen setting through feedback control, while stirring reduces internal temperature gradients. Its finite heat capacity buffers disturbances; external heating and cooling replace the ideal heat bath's unlimited capacity. This is useful for thermometer calibration and for maintaining a blackbody calibration source at a known temperature.

Joel B. Fowler's A Third Generation Water Bath Based Blackbody Source reports a NIST test of a stirred water bath containing a copper blackbody cavity. A bottom plate supplied heating and cooling; a double stirrer circulated water. Electronic feedback control regulated heating. An electronic thermometer with a thermistor probe provided the readout. The preparation was to approach a selected setting, wait for stabilization, adjust against the measured temperature and then log continuously.^[ref]

To test spatial uniformity, one thermistor stayed near the cavity tip while another visited ten locations at three heights. The simultaneous reference distinguished spatial differences from temporal drift. An insulated dome reduced thermal conduction along probe casings. Table 1 reports nonzero departures at several settings: circulation did not make every location identical. The measurement uncertainty analysis separately included uniformity, stability, setting, immersion and readout contributions. These tests supported a bounded temperature assignment around the cavity, rather than infinite heat capacity or perfect thermal equilibrium throughout the apparatus.^[ref]

A concrete experiment turns the phase coexistence curve into pairs of temperature and saturation pressure. A 1969 paper reports water measurements made in 1942 and 1947–1949 using a boiler and precision mercury manometer. Platinum resistance thermometers read the boiler temperature through a resistance bridge. The pressure-transfer line passed through an ice-water condenser and a dry-ice trap; helium filled connecting sections. Corrected mercury column height gave the pressure at the thermometer coils, including hydrostatic pressure contributions from water vapor, helium, mercury, and mercury vapor.^[ref]

Controls included normal/reversed-lead readings, readings at two thermometer electric currents extrapolated to zero current to remove self-heating, repeated bridge calibration, and daily triple point thermometer calibration checks. Cellar temperature measurements corrected fluid-column and gauge-block thermal expansion. One measurement series was excluded because a trapped gas bubble was suspected of corrupting the mercury head. The paper adjusted measurements to round temperatures using previously tabulated slopes; those adjusted values therefore are not independent measurements of those slopes.^[ref]

The thermodynamic inference requires liquid and gas in phase coexistence, a well-defined common temperature, and a pressure corrected to that same location. Leaks, contamination, or a temperature gradient can spoil that interpretation. A separate calorimetry measurement of latent heat, together with the two specific volumes, allows a test of the full Clapeyron equation. Converting a measured phase coexistence curve slope directly into latent heat instead assumes that equation; it is not an independent test.

The Seebeck coefficient $S(T)$ quantifies the thermoelectric voltage response per small temperature difference at zero electric current. Use $S=-dV/dT$ along a homogeneous material, with voltage understood through electrochemical potential. Its units are volts per kelvin, commonly $\mu V/K$. The word thermopower is also used, despite this being a voltage response rather than power; some authors use that word for $|S|$, so check their convention. ^[ref].

For the A-minus-B terminal order in the Seebeck effect experiment,Thus a small temperature difference around $T_0$ gives $V_{AB}\approx [S_A(T_0)-S_B(T_0)]\Delta T$. The finite-interval ratio $V_{AB}/\Delta T$ is an average of the coefficient difference, not necessarily its value at either endpoint. Ordinary two-material measurements determine a relative Seebeck coefficient: to recover $S_A$ from the measured difference, add a separately known $S_B$. Neither the choice of a zero for voltage nor the use of a familiar reference material makes $S_B$ identically zero.

A concrete research example is Martin and colleagues' 2021 certification experiment. They measured small temperature differences and voltages on six silicon–germanium specimens at 13 base temperatures between $295$ and $900\mathrm{K}$. At each setting they acquired 40 voltage-and-temperature-difference pairs and used linear regression with a free intercept, so a constant voltage offset did not become part of the slope. Platinum voltage leads required a reference-material correction; their terminal order gave $S_{\mathrm{measured}}=S_{\mathrm{Pt}}-S_{\mathrm{specimen}}$, hence $S_{\mathrm{specimen}}=S_{\mathrm{Pt}}-S_{\mathrm{measured}}$. This is an actual measurement protocol, unlike the toy plot above. ^[ref].

Microscopically, a temperature gradient changes charge carrier transport until an opposing electrical response balances it at zero net electric current. In the usual simple semiconductor picture, electrons and electron holes give opposite Seebeck coefficient signs, and their contributions are weighted by their electrical conductivities. The Seebeck coefficient is a transport property: density of states alone does not determine it, since carrier motion and scattering also matter. Its usefulness extends beyond temperature sensing to diagnosing which carriers transport electric charge. ^[ref].

The Seebeck effect is the generation of a thermoelectric electromotive force by a temperature difference. In a thermocouple, two different homogeneous electrical conductors A and B connect a measuring junction at $T_h$ to two reference connections at a common $T_c$. With negligible electric current, the measured terminal voltage is determined by the difference of their Seebeck coefficients. A hot junction alone does not specify the reading: the reference temperature, the second material, and the terminal order matter.

For a motivating experiment, join one end of A to one end of B, insulating the wires elsewhere. Connect their free ends to identical meter leads at reference temperature $T_c$, for example in an ice-water heat bath. A meter with input electrical resistance much larger than the thermocouple's electrical resistance approximates an open circuit, meaning $I\approx 0$. Keep the measuring junction in a separately controlled heat bath at $T_h$ and wait for stable junction temperatures. Both reference connections must share $T_c$; identical leads with identical endpoint temperatures have cancelling thermoelectric contributions, allowing the meter to be elsewhere. ^[ref].

Define the reading as $V_{AB}=V_A-V_B$, with the positive input on the A reference terminal and the negative input on B. Use the Seebeck coefficient convention in which the voltage change along a homogeneous wire at zero electric current is $dV=-S(T)dT$. Following the path from A at $T_c$ to the common junction at $T_h$, then back along B to $T_c$, givesThis formula assumes homogeneous materials whose Seebeck coefficients depend only on the local temperature, and no additional uncompensated lead contributions. Here voltage is the electrical transport voltage associated with electrochemical potential, rather than a direct measurement of an isolated equilibrium contact electric potential.

Start with $T_h=T_c$, then increase $T_h$ while holding $T_c$ fixed and record $V_{AB}$ at each stable setting. In the ideal model the equal-temperature reading is zero. If $S_A-S_B>0$ throughout the interval, A reads positive when $T_h>T_c$. Exchanging meter inputs reverses the reading; exchanging the two junction temperatures reverses the oriented integral. Repeating the run with identical homogeneous materials gives zero because $S_A=S_B$. These controls test the stated model; a nonzero experimental baseline should be measured rather than silently removed by assuming it is zero.

For an explicitly toy linear example, stipulate constant Seebeck coefficients $S_A=25\mu V/K$ and $S_B=-15\mu V/K$. Choose reference temperature $T_c=20{}^{\circ }\mathrm{C}$ and measuring-junction temperature $T_h=70{}^{\circ }\mathrm{C}$. A temperature difference has the same numerical size in kelvins and on the degree Celsius scale, soSwapping the inputs gives $-2.00\mathrm{mV}$. Bringing both junction regions to the same temperature gives zero. With temperature-dependent Seebeck coefficients, shifting both junction temperatures equally while keeping their difference fixed can change the reading. No actual material pair is claimed to have the stipulated coefficients over this interval.

Real thermocouples need a calibrated voltage-versus-temperature relation: the constant-Seebeck coefficient approximation is useful only where the coefficient difference changes little. Wire inhomogeneity can make the result depend on where the temperature gradient lies, and reference connections at unequal or unknown temperatures introduce errors. An electronic reference correction uses a measured reference temperature in place of an ice bath. ^[ref]. Burns and colleagues provide the standardized reference functions for common letter-designated thermocouple types; the toy line above is not a replacement for them. ^[ref].

The historical experiment used a different detector. In his 1821–1823 investigations, Seebeck found that a closed electrical circuit of two dissimilar materials with junctions at different temperatures deflected a compass needle. The modern explanation is that thermoelectric electromotive force drives an electric current, whose magnetic field deflects the needle. The open-circuit experiment above detects voltage while drawing negligible electric current. The striking connection is that a thermal imbalance can become an electrical signal and then a magnetic response. ^[ref].

For a saline Stefan problem with common thermal properties, $ϵ=\sqrt{D/\kappa }\ll 1$, and fixed latent-to-sensible heat ratio, set $T_0=-m{C}_0$, ${\theta }_{-}=T_0-T_{-\infty }$ and ${\theta }_{+}=T_{\infty }-T_0>0$. A finite-$\lambda$ similarity solution has $T_i=(T_{\infty }+T_{-\infty })/2+O(ϵ)$. At the onset of constitutional supercooling, $\lambda =O(ϵ)$ and the error in this mean is $O({ϵ}^2)$. The salt-rejection function for a saline Stefan front then gives $m\lambda C_i=({\theta }_{-}-{\theta }_{+})/(2\sqrt{\pi })+O({ϵ}^2)$, whereas $\sqrt{Dt}{T}_{l,x}(a^{+})=ϵ{\theta }_{+}/\sqrt{\pi }+O({ϵ}^2)$. Comparing the actual temperature gradient to the liquidus gradient gives the displayed onset, with supercooling on the side of greater ${\theta }_{-}$. For a freezing front at $a\ge 0$ and $D<\kappa$, the ratio of these two gradients decreases into the liquid, so their interface comparison also decides whether a supercooled interval occurs anywhere ahead. The formula is an asymptotic criterion; finite-$ϵ$ onset requires the full Stefan condition and salt balance.

For $P_0>0$, $\alpha \ne 0$, $T\ge T_0$ and physical $T>0$, the atmospheric pressure-temperature profile $P=P_0\exp (\alpha \sqrt{T-T_0})$ is equivalent toThe branch restriction is essential: squaring does not create a second physical branch. Its temperature gradient is $dT/d\log P=2\sqrt{T-T_0}/\alpha$. Thus, in hydrostatic equilibrium, $\alpha <0$ gives an atmospheric thermal inversion, while $\alpha >0$ gives outward cooling. At $T=T_0$ the inverse branch has zero derivative; $\alpha =0$ gives constant pressure and cannot describe a nontrivial hydrostatic vertical interval.

Stellar convective instability occurs when an adiabatic process displaces a fluid element and buoyancy amplifies the displacement. For uniform composition the Schwarzschild criterion gives instability when the actual temperature gradient exceeds the adiabatic temperature gradient; composition gradients instead require the Ledoux criterion.

Set $\delta =\sqrt{2\kappa /\alpha }$, $R=\alpha h/V$ and $S=L/[c_p(T_0-T_E)]$. In the regular limit $R\gg S\gg 1$, at fixed finite thermal strain number $\alpha h^2/\kappa$ and bounded phase fraction, the heat equation reduces to $\kappa T^{\prime \prime }+\alpha z{T}^{\prime }=0$. Integration gives the displayed error function profile. Matching the roof temperature gradient $G$ yieldsThe product composition is $C_s=C_0-\alpha (T_0-T_E)\delta [1-e^{-(h/\delta {)}^2}]/[mV\sqrt{\pi }\mathrm{erf}(h/\delta )]$. The asymptotic reduction requires admissible phase fractions. If the thermal strain number grows singularly with $R$, the two ordering inequalities alone do not control boundary-layer errors; the full strain-driven steady mush equations retain the latent-heat term.

The temperature gradient is the gradient $\nabla T$ of the temperature field. It points in the direction of fastest temperature increase.

Thermal boundary resistance describes a temperature drop between the two sides of a material interface carrying heat flux. Efficient thermal conduction within each material does not guarantee efficient heat transfer between them. The solid–helium case is Kapitza resistance; the broader concept also includes solid–solid interfaces. ^[ref]

Let $T_s$ and $T_l$ be the temperatures extrapolated to the interface from its solid and liquid sides, each treated as internally in thermal equilibrium. For uniform normal heat flux $q>0$ through area $A$, define the finite-drop, area-specific thermal resistance and the resistance of the whole interface byThe macroscopic jump represents an unresolved interfacial region; it is not a claim that one point has two equilibrium temperatures. At nonzero heat flux, the two sides are outside mutual thermal equilibrium. A measured difference between distant thermometers also contains bulk temperature gradients, which must be measured or bounded before assigning all of it to the interface.

An energy balance explains why a temperature jump can coexist with continuous heat flux. For a stationary interfacial region with no local work or heat source,The first law of thermodynamics therefore requires $q_{\mathrm{in}}=q_{\mathrm{out}}=q$ in a steady state. It does not require $T_s=T_l$. For a heated sample with electrical power $P=VI$, an independently determined outward parasitic loss $P_{\mathrm{leak}}$ and no other inputs,Here $V$ is the heater's electric potential difference and $I$ its electric current. Subtracting a lead loss means subtracting power before dividing by area; $q$ is not the total heating power. Unknown losses and stored internal energy prevent that simple steady inference.

For a one-dimensional comparison, put an interface between two slabs with the same area $A$, positive constant thermal conductivities $k_s,k_l$ and lengths $L_s,L_l$. Neglect side losses. Integrating Fourier's law across each slab givesThus, for a linear interface with constant $r_B$, the series result isThese slab formulas are a chosen thermal conduction model, not a bulk transport law for moving liquid helium. If the interface law is nonlinear, substitute its own relation between $q$ and $T_s-T_l$ into the preceding balance; a single constant resistance cannot describe every load. Increasing bulk thermal conductivity has diminishing benefit when the interfacial drop dominates. Increasing usable contact area reduces the whole-interface thermal resistance if the local interface law and the other conditions remain fixed.

Thermal conduction transfers heat through microscopic interactions without bulk transport of matter. To predict the heating load through a bar, combine Fourier's law with conservation of energy and the imposed end temperatures. In the isotropic model with positive thermal conductivity $k$, the conductive heat flux points from high to low temperature, opposite to the temperature gradient. A spoon warming along its handle and heat escaping through a wall are familiar examples.

Consider a bar of length $L=0.10\mathrm{m}$, constant cross-sectional area $A=0.010{\mathrm{m}}^2$ and constant thermal conductivity $k=2\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. Maintain its ends at $T_h=40{}^{\circ }\mathrm{C}$ and $T_c=20{}^{\circ }\mathrm{C}$, with $x$ increasing toward the colder end. Assume a steady state, one-dimensional thermal conduction, no internal sources and no heat transfer through the sides. The temperature falls by $20\mathrm{K}$ over $0.10\mathrm{m}$, givingThe difference between the two readings in degree Celsius units is the same $20$-kelvin difference. The positive heat flux carries heat toward increasing $x$.

The general solution under these assumptions isA proof outline is enough to explain the straight line: conservation of energy in any small slice, with no accumulation, sources or side losses, gives $d\dot{Q}/dx=0$. Constant cross-sectional area gives $dq/dx=0$; constant thermal conductivity and Fourier's law then give $d^{2}T/d{x}^2=0$. Two integrations and the end boundary conditions determine the displayed affine function. This is why a visually simple temperature profile can encode a continuous flow of energy. ^[ref]

Doubling the length halves the rate, while doubling the cross-sectional area doubles it, at fixed end temperatures and thermal conductivity. This explains why a long narrow support can limit unwanted heat transfer, while a short broad path can help remove heat. Joining materials leads to thermal resistance in series. During initial warming, internal energy can accumulate and the temperature profile need not be a straight line; the time dependence is described by the heat equation and thermal diffusivity.

Now join two layers, each of length $L_1=L_2=0.10\mathrm{m}$, with constant cross-sectional area $A=0.010{\mathrm{m}}^2$ and stipulated constant thermal conductivities $k_1=1$ and $k_2=4\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. Keep the outer ends at $40$ and $20{}^{\circ }\mathrm{C}$. Assume a steady state, one-dimensional thermal conduction, no internal sources or side losses, and perfect contact: the temperature is continuous at the interface, with no extra contact drop. For each layer, use the area-specific thermal resistance $r_i=L_i/k_i$ and $R_i=r_i/A$. The series calculation gives

The heat flux is continuous: otherwise energy would accumulate at the interface. Yet Fourier's law gives different temperature gradients, $dT/dx=-q/k_1=-160\mathrm{K}{\mathrm{m}}^{-1}$ and $dT/dx=-q/k_2=-40\mathrm{K}{\mathrm{m}}^{-1}$. Lower thermal conductivity requires a steeper temperature gradient to carry the same heat flux; equal lengths therefore give drops in the ratio $4:1$. A kink in the temperature profile records the change of material while perfect contact keeps the temperature itself continuous.

Thermal conductivity is the material response coefficient in Fourier's law. For an isotropic material it is a scalar $k$, with units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$; larger positive $k$ means a larger conductive heat flux for the same temperature gradient. It can depend on temperature and composition. Direction-dependent response in an anisotropic material is described by a thermal conductivity tensor. The two worked examples use stipulated positive constants, rather than measured values for named materials.

Thermal conductivity describes material response, whereas thermal resistance also includes geometry: for the steady bar, $R_{\mathrm{th}}=L/(kA)$. Doubling $k$ at fixed length, cross-sectional area and end temperatures halves thermal resistance and doubles the rate of heat transfer. Doubling length changes thermal resistance without changing the material's thermal conductivity. This distinction lets a designer improve cooling either by choosing a higher thermal conductivity or by shortening and widening the path. It also explains why the less conductive layer in thermal resistance takes the greater temperature drop.

Thermal diffusivity answers a different question: how rapidly a temperature disturbance spreads. Its usual model $\kappa =k/(\rho c_p)$ also involves mass density $\rho$ and specific heat capacity $c_p$, so equal thermal conductivities need not imply equal rates of warming.

The thermal conductivity tensor maps a temperature gradient to the negative heat-flux density. In passive reciprocal conduction it is symmetric positive definite. This permits direction-dependent transport while ensuring positive dissipated energy.

With thermal diffusivity $\xi$, a Fourier mode of angular wavenumber $k$ decays at rate $\beta =\xi k^2$. To avoid confusing this $k$ with thermal conductivity, use $\alpha =\xi$ and angular wavenumber $K$ below: the same rate is $\beta =\alpha K^2$, with units ${\mathrm{s}}^{-1}$.

Consider an insulated bar $0\le x\le L$ with constant thermal diffusivity, no sources and no heat flux through either end or the sides. An original initial temperature profile and its exact evolution areThe cosine has zero temperature gradient at both ends, satisfying the insulated Neumann boundary conditions. Its spatial mean is zero, so conservation of energy leaves the mean temperature $\overline{T}$ unchanged while the nonuniform part disappears. A short derivation substitutes $T=\overline{T}+A(t)\cos (K_{n}x)$ into the heat equation: the second spatial derivative supplies $-K_n^2$, hence $A^{\prime }=-\alpha K_n^{2}A$. The exponential decay time is ${\tau }_n=1/(\alpha K_n^2)$; doubling angular wavenumber quarters this time. If cycles-per-length spatial frequency $f_x$ is used instead, $K=2\pi f_x$ and the rate is $4{\pi }^2\alpha f_x^2$.

For the two model materials in thermal diffusivity and $L=0.010\mathrm{m}$, ${\tau }_{1,A}\approx 5.07\mathrm{s}$ and ${\tau }_{1,B}\approx 20.3\mathrm{s}$. After $10\mathrm{s}$, the remaining amplitude fractions are approximately $0.139$ and $0.610$. A narrow hot-and-cold pattern disappears faster than a broad one because its temperature curvature, and hence the imbalance in conductive heat flux, is larger.

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.