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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 conductance is the heat transfer rate divided by the driving temperature difference in a linear contact model. It is the reciprocal of thermal resistance, with units $\mathrm{W}{\mathrm{K}}^{-1}$; thus $\dot{Q}=G(T_1-T_2)$ for heat transferred from body 1 to body 2.

The effective rate of heat transfer per body-to-environment temperature difference. Insulation, posture and perfusion can change it.

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.

A thermal diffusion wave is a periodically forced solution of the heat equation, with spatially decreasing amplitude and varying wave phase. In a semi-infinite rod $x\ge 0$, assume constant thermal diffusivity, negligible side losses, a nearly uniform temperature across each cross-section, and the boundary conditions $\theta (0,t)={\Theta }_0\cos \omega t$ and $\theta (x,t)\to 0$ as $x\to \infty$. A steady periodic thermal state then satisfiesChoose the square root in the complex numbers with $a>0$ so that the amplitude decays into the rod. Substitution of the phasor into the heat equation proves the formula: $i\omega \hat{\theta }=\alpha {\hat{\theta }}^{\prime \prime }$. Taking the real part gives the displayed cosine. In this convention $a$ is the spatial attenuation slope and $b$ is the spatial lag slope, both in ${\mathrm{m}}^{-1}$; $\gamma$ combines them into one complex number. If a complex wavenumber $K$ is defined by $e^{i\omega t-iKx}$ instead, $K=b-ia$.

With $\Delta x=x_2-x_1>0$, let $A_j=|\hat{\theta }(x_j)|$ and define a positive downstream lag $\Delta \varphi$. Thenand either measurement determines thermal diffusivity:The logarithm uses an amplitude ratio, not a ratio of squared amplitudes. The lag is in radians and must be tracked continuously across cycles: a measured phase difference modulo $2\pi$ leaves an integer-cycle ambiguity. Use closer sensor spacings or vary frequency continuously to resolve that ambiguity. Equal attenuation and lag slopes provide a useful diagnostic of the no-side-loss model.

The apparent phase velocity is $\omega /b=\sqrt{2\alpha \omega }$, which depends on frequency. A thermal diffusion wave therefore does not give a material-independent travel speed for heat; it is a forced pattern in a dissipative diffusion equation. The ideal heat equation still has infinite propagation speed, and the observed lag is not the arrival time of a sharp front.

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.

Two internally equilibrated systems are in thermal equilibrium if bringing them into contact that permits only heat exchange causes no net heat transfer. In ordinary nonrelativistic thermodynamics without appreciable gravitational temperature shifts, their thermodynamic temperatures are equal. Merely preventing heat transfer with insulation does not establish thermal equilibrium: two insulated bodies can have different temperatures. ^[ref]

A thermometer illustrates the definition. Its own temperature settles to the object's temperature after thermal contact, provided its effect on the object is negligible. The existing zeroth law of thermodynamics makes this comparison transitive: two objects separately in thermal equilibrium with the same thermometer are in thermal equilibrium with each other. This lets one calibrated scale compare different materials.

Equal temperatures do not imply equal internal energies or equal heat capacities. Nor does thermal equilibrium by itself imply mechanical equilibrium or chemical equilibrium; these require additional allowed exchanges. There is also a distinction between no net exchange in thermal equilibrium and a constant temperature maintained by equal incoming and outgoing heat. The latter can be a steady state with ongoing entropy production. Microscopic exchanges can continue at thermal equilibrium, but their opposing contributions balance on average.

Historically, OpenStax credits the name “zeroth law” to Ralph Fowler in the 1930s: the already numbered laws needed a logically earlier statement about thermal equilibrium. ^[ref]

Thermal loading by a contact temperature sensor is the disturbance of an object's temperature caused by exchanging heat with the temperature sensor used to measure it. A sensor initially colder than a small object draws heat from the object: the final reading can be accurate for the disturbed system while differing from the object's original temperature.

An illuminating idealization treats the object and sensor as two uniform bodies with constant heat capacities $C_o$ and $C_s$, joined by thermal conductance $G$ and isolated from everything else. Their energy conservation equations areAdding the equations shows that $C_o{T}_o+C_s{T}_s$ is constant. Consequently,For $C_o=0.20J/K$, $C_s=0.020J/K$, $T_{o0}=80{}^{\circ }\mathrm{C}$ and $T_{s0}=20{}^{\circ }\mathrm{C}$, the common equilibrium is $74.55{}^{\circ }\mathrm{C}$. The measurement has cooled the object by about $5.45\mathrm{K}$.

Subtracting the two differential equations gives an exponentially decaying temperature difference with time constantIncreasing the contact thermal conductance makes the disturbance happen faster without changing this isolated model's final equilibrium. Reducing the sensor heat capacity reduces the disturbance. A real object also exchanges heat with its surroundings, while sensor leads can provide another heat transfer path.

An electrically read sensor can additionally generate Joule heating. That is a separate loading mechanism: the thermometry guide measures resistance-thermometer self-heating by changing the measuring electric current and extrapolating toward zero dissipated electric power.^[ref]

Thermoacoustics studies coupling between thermal processes and acoustic motion. Oscillatory compression changes gas temperature; heat exchange at the right phase can amplify a sound wave or let sound transport heat.

In an engine, a maintained temperature difference can drive acoustic oscillations. In a refrigerator, externally supplied acoustic work can pump heat. The timing and location of heat exchange determine the energy transfer, so a temperature gradient alone does not guarantee self-sustained sound. Viscous and thermal boundary layers are important near the heat-exchanging structures.

A thermodynamic cycle returns a working substance to its initial thermodynamic state. Every state function, including internal energy, has zero net change over the complete cycle. With heat entering and work done by the substance taken positive, the first law of thermodynamics gives $W_{\mathrm{net}}=Q_{\mathrm{net}}$.

For a quasistatic simple compressible system, $W_{\mathrm{net}}=\oint pdV$ is the signed area enclosed in its pressure-volume diagram. Clockwise traversal gives positive net expansion work in the usual axes, while reversed traversal changes the sign. A heat engine accepts heat at higher temperature and rejects some at lower temperature; a refrigerator uses work to reverse that transfer. Returning the working substance to its initial state does not require its reservoirs or surroundings to return to theirs. A cycle can therefore produce net entropy even though the substance's net entropy change is zero.

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.

A free energy is a thermodynamic potential that compares possible states under specified environmental constraints. For a closed system of fixed composition, treated as a simple compressible system with only pressure-volume work, stable thermodynamic equilibrium minimizes Helmholtz free energy at fixed temperature and volume, or Gibbs free energy at fixed temperature and pressure. The choice follows the experiment: a rigid sealed vessel and a sealed cylinder with a freely moving loaded piston impose different constraints. Both may exchange heat with a thermal reservoir. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19.

The useful insight is that internal energy alone cannot rank these alternatives: exchanging heat also changes the surroundings' entropy. Subtracting $TS$ packages that accounting into a property of the chosen system. Adding $pV$ additionally accounts for displacing the surroundings at fixed pressure. A free energy is a state function; supplied heat describes a transfer during a process. Under the stated constraints, supplied heat is $Q=\Delta U$ at fixed volume or $Q=\Delta H$ at fixed pressure, whereas neither $\Delta F$ nor $\Delta G$ generally equals $Q$.

A lower free energy identifies a thermodynamic preference, without specifying a timescale. A barrier can trap a system in a local minimum, producing metastability although a lower global minimum exists. The enzyme example illustrates the corresponding distinction for chemical reactions: changing the route can change the reaction rate while preserving the endpoint Gibbs free energy difference and chemical equilibrium.

A thermodynamic process changes a thermodynamic system between states. A quasistatic process supplies equilibrium intermediate states and therefore a path on equilibrium coordinates. An irreversible thermodynamic process can have equilibrium endpoints while lacking a uniform intermediate temperature or pressure. Endpoint values still determine changes in a state function; computing heat and work requires the actual interaction history.

Thermodynamics studies energy transfers and the constraints on macroscopic changes. Its vocabulary connects internal energy, entropy, temperature, pressure and volume to heat and work. The first law of thermodynamics balances energy; the Second law of thermodynamics decides which changes can occur spontaneously and limits how much work a heat engine can deliver. These principles let us bound performance before choosing a material or a detailed machine design. Applications include power generation, propulsion and refrigerators. ^[ref]

Start by choosing a system boundary. A closed system can exchange energy but no matter; an isolated system exchanges neither. A state function, such as internal energy or entropy, depends on the equilibrium state. A process specifies how that state changes. Thus a body's temperature is a property it has, whereas heat records energy crossing its boundary during a process. A body contains internal energy, rather than a stored quantity of heat.

Two insulated bodies initially at different positive temperatures approach thermal equilibrium when brought into thermal contact. Their combined energy is conserved, but their combined entropy increases. They do not spontaneously separate again into a hotter body and a colder one. This familiar asymmetry makes thermodynamics interesting: conserving energy leaves many imaginable changes available, and entropy supplies a further constraint. The same distinction underlies both entropy maximization under thermal contact and the thermal efficiency limit in Carnot's theorem.

A thermodynamic system is the part of the physical world selected for an energy and matter balance. Draw its boundary before classifying an interaction as heat, work or transported matter. The boundary may move: a sealed piston encloses a closed system even while its volume changes. An isolated system additionally excludes all external energy transfers.^[ref]

Boundary choice changes the ledger. If an electrical heater warms water, choosing only the water counts incoming heat; including the heater counts electrical work entering through its leads. The combined internal energy records storage, without separate stored categories for the two transfer mechanisms.^[ref]

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.

Thermoelectric cooling uses the Peltier effect to remove heat from a load using electrical work. A typical thermoelectric module connects many pairs of different semiconductors electrically in a series circuit and thermally in parallel between two faces. For an idealized pair with $S_p>S_n$, positive conventional current crosses n to p at the cold bridge and p to n at the hot return. With common bridge material and isothermal contacts, the intervening metal contributions telescope: the net pair coefficient is $S_p-S_n$. For $N$ identical pairs define the module Seebeck coefficient $\alpha =N(S_p-S_n)>0$.

Choose fixed face temperatures $T_h\ge T_c>0$, $\Delta T=T_h-T_c$, module electrical resistance $R>0$ and open-circuit thermal conductance $K>0$. All three module parameters $\alpha ,R,K$ are constants over the chosen operating range. Positive $I$ is the orientation absorbing Peltier effect heat at c. Define $Q_c$ as net heat extracted from the cold reservoir per unit time, $Q_h$ as net heat delivered to the hot reservoir per unit time, and $P_{\mathrm{in}}$ as electrical power entering the module. Despite the conventional letters $Q_c,Q_h$, both denote power in watts, not accumulated energy. Ferrotec gives the corresponding single-stage equations and distinguishes module performance from bypass leaks.^[ref]

At fixed face temperatures, differentiation of the thermoelectric cooler energy balance givesThus current chosen for maximum cooling power is not automatically current for maximum coefficient of performance. Cooling exists only when $(\alpha T_c{)}^2>2RK\Delta T$, and then only between the two positive rootsAt equality the parabola merely touches $Q_c=0$. For the toy module the interval is approximately $0.767<I<10.433\mathrm{A}$, with $I_{\ast }=5.60\mathrm{A}$ and $Q_{c,\max }=23.36\mathrm{W}$. At $12\mathrm{A}$, the object is heated even in the nominal cooling polarity: $Q_c=-17.6\mathrm{W}$. This mathematical interval supplies no permission to exceed an actual device's current or temperature ratings.

The fixed-face constraint matters. With fixed $T_h$ but a freely cooling object, $T_c$ must be solved from its heat load and leaks rather than held at $280\mathrm{K}$ while claiming an arbitrary temperature drop. For zero load and optimal current, the same model givesHere $Z$ has units ${\mathrm{K}}^{-1}$ and is a module parameter. The toy model gives $Z=0.0020{\mathrm{K}}^{-1}$ and $T_{c,\min }\approx 241.62\mathrm{K}$ when $T_h=300\mathrm{K}$. This is an ideal zero-load limit with stipulated constant properties; cooling a real load requires $Q_c$ above zero and reduces the attainable drop. Contact thermal resistance, a warming heat sink, external leaks and temperature-dependent coefficients further change the result. Electrical control is convenient, but conservation of energy still requires a thermal path that removes the extracted heat plus the electrical input.