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With the Seebeck coefficient convention $dV=-SdT$ at zero electric current, and the transported-heat convention of Peltier coefficient, the Kelvin thermoelectric relation is$T$ is absolute temperature in kelvins; use consistent material references. This scalar linear response relation assumes local thermal equilibrium and reciprocal transport without time-reversal-breaking fields or order. Onsager reciprocal relations supply the symmetry; conservation of energy alone is insufficient. A magnetic field requires a reversed-field reciprocity comparison.^[ref]

In particular, the A-minus-B thermocouple coefficient in Seebeck effect is $S_A-S_B$, whereas absorbed junction heat for conventional current A to B uses $S_B-S_A$. There is no contradiction: terminal voltage order and crossing direction are separate choices. Multiplying a measured voltage by an absolute temperature without specifying these choices does not determine junction heating.

A short reciprocity calculation makes the factor $T$ visible. At a common reference temperature, write the local linear response with electrical driving field $\mathcal{E}=-\nabla V$ and heat flux $q$ asAt $j=0$, the Seebeck coefficient convention gives $S=L_{12}/(T{L}_{11})$. At $\nabla T=0$, the Peltier coefficient is $q/j=L_{21}/L_{11}$. Thus $L_{21}=L_{12}$ gives $\Pi =TS$. The chosen forces are conjugate to the fluxes in the local entropy production, $j\mathcal{E}/T+q(-\nabla T)/T^2$; arbitrary rescaling of just one force would spoil this argument.

The relation is local. Varying $S(T)$ permits distributed Thomson effect heating, with $\tau =TdS/dT$; the constant-coefficient model below omits it. Endpoint substitutions into a constant-property formula can therefore break its energy accounting.

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.