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A slender planar plume in a saturated porous medium has buoyancy velocity $w=\beta (T-T_0)$, with $\beta ={\rho }_{0}g\alpha \Pi /\mu$. The leading heat equation retains both advection components and lateral thermal conduction: $u{T}_x+w{T}_z=\kappa T_{xx}$. Its convective heat input per span is conserved, $F_h=C_h\int w(T-T_0)dx$. Defining $Q=\beta F_h/C_h$ gives width proportional to $z^{2/3}$ and vertical speed/temperature excess proportional to $z^{-1/3}$.

A growing ice-filled sphere expels liquid because ice is less dense than water. With porosity $\varphi$, $ϵ=1-{\rho }_s/\rho$, and pressure-induced freezing-point depression $T_i=T_m-\gamma p_i$, the Darcy velocity gives $p_i=\mu \varphi ϵa\dot{a}/\Pi$. Quasi-steady spherical thermal conduction gives $\varphi {\rho }_{s}La\dot{a}=k(T_i-T_{\infty })$. Hence $a\dot{a}=k(T_m-T_{\infty })/[\varphi {\rho }_{s}L(1+\Lambda )]$, where $\Lambda =k\gamma \mu ϵ/(\Pi {\rho }_{s}L)$. The interface approaches the far-field temperature at low permeability of a porous medium, while its pressure tends to a finite equilibrium value.

A radiometer thermal link is the intentional path for thermal conduction from receiver to reference block. Near a chosen operating point, write $Q_{\mathrm{link}}\simeq G(T-T_b)$ with local conductance $G$ in $\mathrm{W}{\mathrm{K}}^{-1}$. A weaker link increases the steady temperature rise per added power, $\delta T\simeq \delta P/G$, but also increases the uncontrolled relaxation time $\tau =C/G$ when the link dominates other heat transfer. These are local linear approximations: actual thermal conductivity depends on temperature, and active feedback control introduces its own response. NIST explicitly describes the tradeoff between weaker links and smaller receiver heat capacity.^[ref]

Loss of summer sea ice need not be irreversible at fixed external forcing. Open water can release excess heat in autumn, and thin ice grows faster than thick ice during winter because it offers less resistance to thermal conduction. Continued climate change can nevertheless shift the long-term ice cover downward. A one-summer perturbation and persistent forcing are different physical experiments.

A probe's sheath and leads carry heat between the sensing region and the surroundings through thermal conduction. In a stable, sufficiently uniform heat bath, compare the indicated temperature as insertion depth increases, allowing settling each time. A plateau supports adequate immersion for that apparatus and required uncertainty; it does not prove that every depth or medium works. Keep the sensing element and reference thermometer at comparable depths, and check bath gradients independently, so moving the probe does not merely sample a different temperature. The documented platinum resistance thermometer comparison calibration includes these controls.

Sea ice is frozen ocean water. A local thermodynamic description uses thermal conduction, a surface energy balance and the Stefan condition; salinity, mechanical transport and fractional ice coverage add further physics.

A single-category model represents local ice by one thickness $h$ and one surface temperature $T_s$. With basal freezing temperature $T_m$, atmospheric temperature $T_A$, thermal conductivity $k$ and atmospheric heat transfer coefficient ${\lambda }_A$, quasistatic thermal conduction giveswhile $T_s<T_m$. If $F_O$ is upward ocean heat flux, the Stefan condition gives $\rho L\dot{h}={\lambda }_A(T_m-T_A)/(1+{\lambda }_{A}h/k)-F_O$. During surface melting, $T_s=T_m$ and $\rho L\dot{h}=-{\lambda }_A(T_A-T_m)-F_O$ when $T_A\ge T_m$. Thickness is constrained to be nonnegative, and the open-water energy balance applies after complete melt.

A spherical solid of radius $a$ growing inside a fixed cavity of radius $R$ expels volume at rate $4\pi (1-{\rho }_s/{\rho }_l)a^2\dot{a}$. The Darcy flux outside the cavity gives $p-p_m=\mu (1-{\rho }_s/{\rho }_l)a^2\dot{a}/(\Pi R)$. The common-pressure melting-point slope and quasi-steady spherical thermal conduction give ${\rho }_{s}La\dot{a}=k(T_i-T_{\infty })$. Define $x=a/R$, $\tau =tk(T_m-T_{\infty })/(R^2{\rho }_{s}L)$ and $K=L^2\Pi /[k{T}_m\mu (1/{\rho }_s-1/{\rho }_l{)}^2]$. Elimination yields $(x+x^2/K)x_{\tau }=1$ and the displayed zero-seed integral. The pressure rise is $p^{\ast }x/(x+K)$, with $p^{\ast }=L(T_m-T_{\infty })/[T_m(1/{\rho }_s-1/{\rho }_l)]$. Easy drainage gives square-root growth; low permeability gives an initial square-root regime followed by cube-root growth and pressure saturation. The model stops at $x=1$ and uses a uniform exterior conductivity.

The Stefan–Boltzmann law gives the total hemispherical radiant exitance of an ideal blackbody at absolute temperature $T$:Here $\sigma$ is the Stefan-Boltzmann constant. Total means integrated over all wavelengths; hemispherical means integrated over all outward directions with the projected area weighting. Thus $M_b$ is emitted power per actual emitting area, rather than directional radiance or spectral radiance. Absolute temperature must be in kelvin; raising a Celsius reading to the fourth power gives the wrong result. A uniform surface of area $A$ emits radiant flux $P_{\mathrm{emit}}=A{M}_b$. A real surface instead has $M=ϵ(T)\sigma T^4$, where $ϵ(T)$ is its total hemispherical emissivity. The older symbol $F$ in the formula above denotes this emitted radiant exitance.^[ref]

For an ideal blackbody, doubling absolute temperature multiplies radiant exitance by $16$. With the approximate Stefan-Boltzmann constant $5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, the radiant exitances at $300\mathrm{K}$ and $600\mathrm{K}$ are approximately $459.300$ and $7348.805\mathrm{W}{\mathrm{m}}^{-2}$. This steep response explains why thermal radiation becomes so consequential at high temperature. It also enables heat transfer through a vacuum, where thermal conduction through a gas and convection are absent.

A short derivation integrates Planck's law in its frequency form. For a blackbody, direction-independent spectral radiance is $B_{\nu }(T)=2h{\nu }^3/[c^2(e^{h\nu /(k_{B}T)}-1)]$, where $h$ is the Planck constant, $k_B$ the Boltzmann constant and $c$ the speed of light. The outward hemisphere contributes $\int \cos \theta d\Omega =\pi$, soExpanding $(e^x-1{)}^{-1}={\sum }_{n=1}^{\infty }{e}^{-nx}$, nonnegative terms allow termwise integration. Repeated integration by parts gives ${\int }_0^{\infty }{x}^3{e}^{-nx}dx=6/n^4$; the Riemann zeta function value $\zeta (4)={\pi }^4/90$ gives the remaining integral as ${\pi }^4/15$. Hence $M_b=\sigma T^4$ with the coefficient in Stefan-Boltzmann constant. The angular integral is essential: omitting it confuses radiance with radiant exitance.

A thermal boundary layer is a thin region in which temperature changes rapidly and thermal conduction remains important. Its thickness can be estimated by balancing thermal diffusion against advection or by requiring a local Rayleigh number to be near its critical value.

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 convection is fluid motion driven by temperature-dependent buoyancy. It transports heat by bulk motion in addition to thermal conduction.

The thermal diffusion length $\ell (t)=\sqrt{\alpha t}$ is the characteristic length over which thermal conduction changes a temperature distribution during time $t$ in the constant-thermal diffusivity model. It is a scale, not a sharp front or distance travelled at a constant speed: the heat equation has infinite propagation speed mathematically. Different definitions of penetration depth introduce numerical factors. In particular, the normalized one-dimensional heat kernel has spatial variance $2\alpha t$, so its root-mean-square width is $\sqrt{2\alpha t}$.

At $t=10\mathrm{s}$, the model materials have thermal diffusion lengths $4.47$ and $2.24\mathrm{mm}$. For $L=10\mathrm{mm}$ their thermal diffusion times are $50$ and $200\mathrm{s}$. These values describe the same heat equation as the thermal diffusion rate of a Fourier mode; the different numerical factors come from the definition of the scale.

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.

For a sharp thermal front moving at fraction $\Gamma$ of the Darcy velocity $U$, a cold fluid of viscosity $\mu$ displacing its hot counterpart of viscosity $b\mu$ has the displayed growth rate in the local, unbounded planar approximation without thermal conduction. The harmonic pressure perturbations decay on either side. Pressure continuity and continuity of perturbed Darcy flux give the velocity response $\delta U=U|a|(b-1)\eta /(b+1)$; the thermal front's kinematic law is ${\eta }_t=\Gamma \delta U$. The prefactor is the thermal-front speed, not the material pore velocity. Increasing viscosity with heating therefore makes the trailing thermal interface unstable even when the leading displacement of the original fluid is stable.

Thermal resistance measures the temperature drop required to drive a given rate of heat transfer. For the steady one-dimensional bar model in thermal conduction,Here $r$ relates the temperature drop to heat flux, while $R_{\mathrm{th}}$ relates it to the rate of heat transfer. The single bar above has $r=0.050{\mathrm{m}}^2\mathrm{K}{\mathrm{W}}^{-1}$ and $R_{\mathrm{th}}=5\mathrm{K}{\mathrm{W}}^{-1}$, so its $20\mathrm{K}$ drop drives $4\mathrm{W}$.

To see why series thermal resistances add, apply Fourier's law in each layer to obtain $\Delta T_i=\dot{Q}{R}_i$. With no accumulation or side losses the rate of heat transfer is the same through both, so adding the drops gives $T_h-T_c=\dot{Q}(R_1+R_2)$. This lets us size layered walls without solving a new differential equation for every layer. The structure matches Ohm's law: temperature difference plays the role of electric potential difference, the rate of heat transfer plays the role of electric current, and thermal resistance plays the role of electrical resistance. ^[ref]

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

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

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

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

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