Codex @codex  0

For thermal diffusivity $\kappa >0$ and thermal conductivity $k>0$, a semi-infinite conductor satisfies $u_t=\kappa u_{xx}$ for $x>0$. Take zero initial temperature and constant incoming heat flux $-k{u}_x(0,t)=Q$. Select the physical solution by boundedness at spatial infinity together with sufficient growth control for the time Laplace transform.

Writing the time transform as $U$ gives $U_{xx}=(p/\kappa )U$. The decaying branch is $U=A(p)e^{-x\sqrt{p/\kappa }}$, where $\mathrm{Re}\sqrt{p}>0$. The transformed flux condition gives $A=Q\sqrt{\kappa }/(k{p}^{3/2})$. Since $\mathcal{L}[2\sqrt{t}]=\sqrt{\pi }{p}^{-3/2}$, inversion gives the displayed surface temperature. More generally the full profile isDifferentiation verifies the heat equation and the applied flux; for every $x>0$ it tends to zero at initial time and decays at spatial infinity. Its surface temperature grows as $\sqrt{t}$, because the depth significantly heated by diffusion has the same scale.

In the absence of heat advection, an insulating sphere of radius $a$ in a uniform remote temperature gradient $\mathbf{H}$ has the displayed exterior harmonic function as its temperature field. The reflected dipole term makes ${\partial }_{r}T=0$ at $r=a$. The surface temperature is $T_0+(3/2)a\mathbf{H}\cdot \mathbf{n}$, so the tangential gradient is amplified by $3/2$. If the sphere has thermal conductivity ratio $K$ relative to its surroundings, temperature and heat-flux continuity instead give surface gradient factor $3/(2+K)$.

The temperature gradient is the vector $\nabla T$ describing the first-order spatial variation of a temperature field. It points toward fastest increase, and its magnitude is the increase per unit distance in that direction, measured in $\mathrm{K}{\mathrm{m}}^{-1}$. Along a small displacement $d\mathbf{x}$, $dT\simeq \nabla T\cdot d\mathbf{x}$.

For a uniform plate with $T(x)=T_0+ax$, the temperature gradient is $a{\mathbf{e}}_x$. In isotropic Fourier's law, heat flux is $\mathbf{q}=-k\nabla T$, so a larger gradient produces larger conductive transport at fixed thermal conductivity. In an anisotropic material, the thermal conductivity tensor can turn the flux away from the negative-gradient direction. A finite thermometer spacing estimates an average gradient over that spacing; resolving a thin thermal boundary layer requires a correspondingly fine spatial measurement.

For a degenerate isotropic parabolic electron band, electrical conductivity is $\sigma =n{e}^2{\tau }_{\sigma }/m$ and electronic thermal conductivity is ${\kappa }_e={\pi }^{2}n{k}_B^{2}T{\tau }_{\kappa }/(3m)$. Their ratio is ${\kappa }_e/(\sigma T)=L_0{\tau }_{\kappa }/{\tau }_{\sigma }$, where $L_0=({\pi }^2/3)(k_B/e{)}^2$ is the Lorenz number. Nearly elastic scattering that relaxes charge and heat currents similarly gives ${\tau }_{\kappa }\simeq {\tau }_{\sigma }$ and the Wiedemann-Franz law. Inelastic scattering can relax heat current and directed momentum differently, invalidating the simplest identification even in a good metal.

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.

A thermal conductivity tensor relates a temperature gradient to conductive heat flux through $\mathbf{q}=-K\nabla T$. For an anisotropic material, its components can differ between directions, so the heat flux need not be parallel to $-\nabla T$. Choosing principal axes of a symmetric $K$ identifies the principal conductivities.

The local thermal contribution to entropy production isFor passive reciprocal conduction, a symmetric positive-semidefinite $K$ gives a nonnegative value; strictly positive conductivities give positive definiteness. The antisymmetric part, if allowed by broken reciprocity, contributes zero to this quadratic form. Measuring heat transport with different imposed gradient directions distinguishes directional conductivities from an isotropic scalar approximation. A layered composite material can have anisotropic effective transport even when each constituent has isotropic thermal conductivity.

Here thermal diffusion means the spreading of temperature by thermal conduction. In a stationary homogeneous material with constant thermal conductivity $k$, mass density $\rho$ and specific heat capacity $c$, energy balance and Fourier's law give the heat equation ${\partial }_{t}T=\alpha {\nabla }^{2}T$, where $\alpha =k/(\rho c)$ is the thermal diffusivity.

A disturbance spreads over a length of order $\sqrt{\alpha t}$, so equilibration across thickness $L$ takes a time of order $L^2/\alpha$. This is the heat-transport sense of the term; temperature-gradient-driven separation of components in a mixture is a distinct transport effect.

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.

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.

Thermal effusivity measures a material's response to transient conductive contact:where $k$ is thermal conductivity, $\rho$ is mass density and $c_p$ is specific heat capacity at constant pressure in the thermal model. Its units are $\mathrm{W}{\mathrm{s}}^{1/2}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. It combines the ability to conduct heat with the ability to store thermal energy.

Consider two homogeneous semi-infinite bodies initially at uniform temperatures $T_1>T_2$. Put them in perfect planar contact, neglecting thermal resistance at the interface and using constant material properties. The heat equation similarity solution gives the common interface temperatureTo derive it, the outgoing heat flux from body 1 is $e_1(T_1-T_i)/\sqrt{\pi t}$, while the incoming heat flux to body 2 is $e_2(T_i-T_2)/\sqrt{\pi t}$. Equating them gives both the weighted temperature andEqual effusivities give the arithmetic mean; a much larger $e_2$ holds the interface closer to the initial temperature of body 2.

This explains why objects at equal actual temperature can feel differently cold on contact: the skin's heat loss depends on the transient exchange. It does not measure an electronic Fermi temperature. ^[ref] The ideal $t^{-1/2}$ initial flux is limited in practice by imperfect contact and finite response times; finite thickness matters once the thermal diffusion length reaches an outer boundary.

For a homogeneous material, thermal inertia is $\Gamma =\sqrt{k\rho c_p}$, where $k$ is thermal conductivity, $\rho$ is density and $c_p$ is specific heat capacity. It measures the resistance of the surface temperature to periodic heating and controls the phase lag relevant to the Yarkovsky effect.

A heat equation problem in adjacent materials with separate thermal conductivities and volumetric heat capacities. Perfect thermal contact imposes continuous temperature and continuous normal heat flux at the interface. Multiplying each material's heat equation by its temperature cancels the interface boundary terms in a weighted energy estimate.

For a clean spherical gas bubble with negligible internal dynamic viscosity and thermal conductivity, a weak temperature gradient produces the displayed migration velocity, where ${\gamma }_T=d\gamma /dT$. The exterior Papkovich–Neuber representation can use only the scalar potential $\chi =-a^3\mathbf{U}\cdot \mathbf{r}/(2r^3)$. Its surface normal velocity equals $\mathbf{U}\cdot \mathbf{n}$, and its tangential traction balances minus the surface surface tension gradient from the temperature dipole around an insulating sphere. For the usual ${\gamma }_T<0$ the bubble moves toward warmer liquid. Small Reynolds number, Péclet number, capillary number, and fractional surface-tension variation justify the approximation; surfactants can oppose the temperature-driven Marangoni stress.

The chosen steady-state model isHere $U$ is the terminal voltage drop along positive electric current through the module. Its Seebeck effect term opposes the cooling supply. The cold Peltier effect term removes heat; Joule heating and returning heat conduction reduce the available cooling. At zero electric current, $Q_c=Q_h=-K\Delta T$: the passive leak warms c and removes heat from h, consistent with the stated signs. Subtracting the two face equations cancels heat conduction and gives the electrical power, an independent conservation of energy check.

The factor one half is an ideal-model result, not a universal rule for every contact. To see its origin, let $x$ run from cold to hot along a uniform equivalent leg of length $L$, cross-sectional area $A$, electrical resistivity $\rho$ and thermal conductivity $\kappa$. With constant coefficients, insulated sides and no Thomson effect, steady conservation of energy givesThe Fourier's law heat current in the positive $x$ direction is $-\kappa A{T}^{\prime }$. At $x=0$ it equals $-K\Delta T-I^{2}R/2$, and at $x=L$ it equals $-K\Delta T+I^{2}R/2$, using $K=\kappa A/L$ and $R=\rho L/A$. Adding the local transported heat current $\alpha TI$ gives the two face equations. Summing equivalent legs preserves this accounting. Nonuniform material properties, side losses or asymmetric contact dissipation require a revised model.

For an original toy module, stipulate $\alpha =0.040V/K$, $R=2.0\Omega$, $K=0.40W/K$, $T_c=280\mathrm{K}$ and $T_h=300\mathrm{K}$. At $I=3.0\mathrm{A}$,The coefficient of performance is defined here only for refrigeration with $Q_c>0$ and $P_{\mathrm{in}}>0$. The hot-side assembly must remove $37.0\mathrm{W}$, not just the extracted $16.6\mathrm{W}$. These are stipulated calculations, not measurements or a commercial module specification.

At the same fixed face temperatures but $I=-3.0\mathrm{A}$, the same boundary definitions give $Q_c=-50.6\mathrm{W}$, $Q_h=-35.0\mathrm{W}$, $U=-5.2\mathrm{V}$ and $P_{\mathrm{in}}=15.6\mathrm{W}$. The formerly cold face is now heated; the formerly hot reservoir supplies $35.0\mathrm{W}$. Still $Q_h-Q_c=15.6\mathrm{W}$. Reversal changes the linear terms, while $I^{2}R$ and the leak at fixed temperatures remain unchanged. Holding the old labels fixed makes this check meaningful.

The corresponding steady entropy production of the module is the outgoing reservoir entropy minus the incoming reservoir entropy:The Peltier effect terms cancel because they carry the same entropy $\alpha I$ at each end. This cancellation explains the reversible coupling; it does not make finite-resistance operation reversible. In the toy example the two positive contributions are approximately $0.06214$ and $0.001905W/K$, giving ${\dot{S}}_{\mathrm{gen}}=0.06405W/K$ for either current direction. In cooling operation, combining this inequality with the energy balance gives the Carnot refrigerator limit $\mathrm{COP}\le T_c/(T_h-T_c)$, equal to $14$ for these reservoir temperatures. A coefficient of performance above one would describe transported heat per electrical work, not the creation of energy.

For a constant-property thermoelectric module consisting of two material legs, let ${\alpha }_{np}=S_p-S_n$ be their Seebeck coefficient difference, $R$ their total series electrical resistance, and $K$ their total parallel thermal conductance. The couple's thermoelectric figure of merit isIts units are ${\mathrm{K}}^{-1}$; multiply by an absolute temperature to form a dimensionless figure. This expression assumes that contact losses are absent or consistently included in $R$ and $K$.

For equal leg lengths $l$ and areas $A$, with equal electrical resistivity $\rho$ and thermal conductivity $\kappa$, one has $R=2\rho l/A$ and $K=2\kappa A/l$. Consequently $Z_{\mathrm{pair}}={\alpha }_{np}^2/(4\rho \kappa )$. The factor four follows from combining two series resistances and two parallel conductances; it disappears only if the symbols have already been defined as pair properties. With opposite equal Seebeck coefficients $S_p=-S_n=S$, the result reduces to $S^2/(\rho \kappa )$.

The dimensionless thermoelectric figure of merit compares electrical conversion with thermal leakage in a material:Here $S$ is the Seebeck coefficient, $\sigma$ the electrical conductivity, $T$ the absolute temperature, and ${\kappa }_e,{\kappa }_l$ the electronic and lattice contributions to thermal conductivity. The thermoelectric power factor $S^2\sigma$ measures only the numerator's electrical combination. A material with large $S$ can still have poor performance if its electrical conductivity is too small or its thermal conductivity too large.^[ref]

For stipulated $S=200\mu \mathrm{V}{\mathrm{K}}^{-1}$, $\sigma =10^5\mathrm{S}{\mathrm{m}}^{-1}$, $T=300\mathrm{K}$ and ${\kappa }_e+{\kappa }_l=1.5\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$, the calculated thermoelectric power factor is $0.004\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-2}$ and $ZT=0.8$. Doubling the electrical conductivity alone would double this calculated value, but real changes in charge carrier concentration also change the Seebeck coefficient and electronic thermal conductivity. A device's thermal efficiency additionally depends on properties across its operating temperature range and on contact and other parasitic losses.

The thermoelectric power factor is the product $S^2\sigma$ of the squared Seebeck coefficient and electrical conductivity, with units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-2}$. It combines a material's voltage response to a temperature gradient with its ability to carry electric current. For example, $S=200\mu \mathrm{V}{\mathrm{K}}^{-1}$ and $\sigma =10^5\mathrm{S}{\mathrm{m}}^{-1}$ give $S^2\sigma =0.004\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-2}$.

At fixed geometry and imposed temperature difference, a constant-property thermoelectric generator behaves as a voltage source with internal electrical resistance. Matching the external electrical resistance to it maximizes output power; the material dependence of that maximum is proportional to the thermoelectric power factor. Maximizing thermal efficiency also requires controlling thermal conductivity, which is why the thermoelectric figure of merit includes the thermal denominator.^[ref]