= Thermoelectric cooler energy balance
{title2=$Q_c=\alpha T_c I-\tfrac12 I^2R-K(T_h-T_c)$}
The chosen steady-state model is
$$
Q_c=\alpha T_c I-\frac12 I^2R-K\Delta T,\qquad
Q_h=\alpha T_h I+\frac12 I^2R-K\Delta T,
$$
$$
U=\alpha\Delta T+IR,\qquad
P_{\rm in}=UI=\alpha\Delta T I+I^2R,\qquad
Q_h-Q_c=P_{\rm in}.
$$
Here $U$ 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> gives
$$
\kappa A T''=-\frac{I^2\rho}{A},\qquad
T(x)=T_c+\frac{\Delta T}{L}x+\frac{I^2\rho}{2\kappa A^2}x(L-x).
$$
The <Fourier's law> <heat current> in the positive $x$ direction is $-\kappa A T'$. At $x=0$ it equals $-K\Delta T-I^2R/2$, and at $x=L$ it equals $-K\Delta T+I^2R/2$, using $K=\kappa A/L$ and $R=\rho L/A$. Adding the local transported <heat current> $\alpha T I$ 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 $\alpha=0.040\,\mathrm{V/K}$, $R=2.0\,\Omega$, $K=0.40\,\mathrm{W/K}$, $T_c=280\,\mathrm K$ and $T_h=300\,\mathrm K$. At $I=3.0\,\mathrm A$,
$$
Q_c=33.6-9.0-8.0=16.6\,\mathrm W,\qquad
Q_h=36.0+9.0-8.0=37.0\,\mathrm W,
$$
$$
U=0.8+6.0=6.8\,\mathrm V,\qquad P_{\rm in}=20.4\,\mathrm W,
\qquad \mathrm{COP}=\frac{Q_c}{P_{\rm in}}\approx0.814.
$$
The <coefficient of performance> is defined here only for refrigeration with $Q_c>0$ and $P_{\rm in}>0$. 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 $P_{\rm in}=15.6\,\mathrm W$. 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 $I^2R$ 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>:
$$
\dot S_{\rm gen}=\frac{Q_h}{T_h}-\frac{Q_c}{T_c}
=\frac{I^2R}{2}\left(\frac1{T_h}+\frac1{T_c}\right)
+\frac{K(\Delta T)^2}{T_hT_c}\ge0.
$$
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 $0.001905\,\mathrm{W/K}$, giving $\dot S_{\rm gen}=0.06405\,\mathrm{W/K}$ 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>.
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