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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.

In a metal, mobile Electrons transport heat as well as electric charge. A temperature gradient creates an imbalance between carriers arriving from warmer and colder regions. The resulting heat current obeys Fourier's law ${\mathbf{j}}_Q=-\kappa \nabla T$ for an isotropic material, where thermal conductivity $\kappa$ has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a bar of area $A$ carrying steady heat flow $P$, $P=-\kappa AdT/dx$ when parasitic losses are negligible.

A kinetic estimate for the electronic contribution iswhere $C_e$ is the electronic heat capacity per volume, $v_F$ the Fermi velocity, and ${\ell }_{\kappa }=v_F{\tau }_{\kappa }$ the heat-transport mean free path. Fermi-Dirac statistics and the Pauli exclusion principle restrict thermal rearrangements to states near the Fermi surface, rather than allowing every Electron to contribute a classical heat capacity. A degenerate parabolic band givesUsing the Drude model form $\sigma =n{e}^2{\tau }_{\sigma }/m$ for electrical transport, the relation between thermal and electrical relaxation times in metals becomesFor scattering that relaxes heat and charge currents similarly, ${\tau }_{\kappa }\simeq {\tau }_{\sigma }$ and the Wiedemann-Franz law follows:Thus good electrical conductors are usually good thermal conductors. The correct quantum Lorenz number differs from a naive classical electron-gas estimate.

Scattering from impurities, boundaries and phonons limits the heat-transport path. In an impurity-dominated low-temperature regime with approximately constant relaxation time, ${\kappa }_e$ is approximately proportional to $T$. Where phonon-limited electrical resistivity is proportional to $T$ and the Wiedemann-Franz approximation is applicable, ${\kappa }_e$ can be roughly temperature independent. Inelastic processes can relax heat and charge currents differently, so the Lorenz ratio is not universal at all temperatures. Lattice phonons add a separate positive thermal contribution; the measured total is $\kappa ={\kappa }_e+{\kappa }_{\mathrm{ph}}$, and the electronic Wiedemann-Franz expression should not be applied blindly to that total.

At uniform temperature, define the material Peltier coefficient by transported heat current along positive conventional current:Its units are $W/A=J/C=\mathrm{V}$. In one dimension, $q=\Pi j-\kappa dT/dx$, with electric current density $j$ and thermal conductivity $\kappa$. Here heat excludes the electrochemical potential contribution to total energy flux.^[ref]

Now orient positive conventional current $I$ from material A into material B, at a common junction temperature $T$. Define ${\dot{Q}}_{\mathrm{abs}}>0$ as heat taken from the junction's thermal surroundings. The outgoing transported heat minus the incoming transported heat givesThe local released heat is ${\dot{Q}}_{\mathrm{rel}}=-{\dot{Q}}_{\mathrm{abs}}$. Authors defining a junction coefficient by released heat therefore use the opposite sign. Exchanging A and B reverses the coefficient; reversing $I$ at fixed A and B reverses the junction's thermal response. An individual material coefficient and a junction difference are different quantities.

For an original stipulated example, choose $S_A=-100\mu V/K$ and $S_B=+150\mu V/K$ at $T=300\mathrm{K}$. The Kelvin thermoelectric relation gives ${\Pi }_A=-0.030\mathrm{V}$, ${\Pi }_B=+0.045\mathrm{V}$. A positive $2\mathrm{A}$ flowing A to B absorbs $(0.075\mathrm{V})(2\mathrm{A})=0.150\mathrm{W}$. A separate $0.010\Omega$ junction electrical resistance would dissipate $0.040\mathrm{W}$ for either direction. If all that dissipation heats the junction, its net absorbed power is $0.110\mathrm{W}$ forwards and $-0.190\mathrm{W}$ backwards. These numbers illustrate a chosen thermal accounting boundary, not measured material data or a universal allocation of contact heating.

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.

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.