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A thermoelectric effect couples heat transfer and the transport of electric charge: a temperature gradient can generate an electromotive force, while an electric current can carry heat and cause heating or cooling at a junction of different materials. This coupling makes electrical measurement of temperature possible through a thermocouple, and permits conversion of a sustained flow of heat into electrical work. A finite open-circuit voltage does not itself deliver power: electrical power requires an electric current through a load. Thermoelectric conversion therefore involves both electromagnetism and thermodynamics. ^[ref].

Production of heat through metabolic processes.

The thermomechanical effect couples a temperature difference to a pressure difference across a superleak. For the closed-chamber model in the apparatus figure, consider neighboring equilibrium states with small velocities, negligible height difference and pure helium-4. Let ${\mu }_m$ be chemical potential per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. The Gibbs free energy differential per unit mass givesThe superleak allows the superfluid to redistribute until the two sides have equal ${\mu }_m$. Setting $d{\mu }_m=0$ therefore givesThe sign matters: heating one side at initially equal pressure lowers its ${\mu }_m$; inflow raises its pressure until chemical balance is restored. The hotter side has the higher pressure. The integral follows the constant-${\mu }_m$ path; it need not be evaluated at fixed pressure when material properties vary appreciably.

For an original local example, stipulate constant $\rho =150\mathrm{kg}{\mathrm{m}}^{-3}$, $s=100\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $T_h-T_c=0.020\mathrm{K}$. The predicted pressure difference is $300\mathrm{Pa}$. These are teaching assumptions, not helium measurements. This static thermodynamics relation predicts neither the required heater power nor the filling time: those require heat leakage, channel conductance and chamber heat capacity. An outlet would introduce kinetic energy and a flowing balance, so the static $300\mathrm{Pa}$ is not a predicted fountain height.

A thermometer measures temperature through a physical response whose relation to temperature is known. Examples include the length change from thermal expansion, a change in electrical resistance, and the pressure of a gas thermometer. If the response function $X=f(T)$ is an injective function over the working range, its measured value determines $T=f^{-1}(X)$ through the inverse function. Thermometer calibration establishes that relation for the particular instrument and measurement conditions. Two matching readings are meaningful only after calibration to a common scale and with the probes sampling comparable equilibrium states. ^[ref]

For a contact thermometer, the sensing part must reach thermal equilibrium with the region being measured. That region may differ from the rest of the object: an immersed probe near a heater can read a local temperature while the surrounding water remains cooler. A stable reading can also be a steady state set by competing heat transfer paths to the sample and the room. Good contact, adequate immersion, mixing where appropriate, and checks for spatial differences help establish which temperature the reading represents. These are measurement controls inferred from the thermal equilibrium requirement, not a guarantee that every flat trace is equilibrium.

A thermometer also changes the sample it touches. Consider an original toy calculation with an initially uniform sample at $T_o=350\mathrm{K}$, a probe at $T_p=300\mathrm{K}$, and constant whole-body heat capacities $C_o=100\mathrm{J}{\mathrm{K}}^{-1}$ and $C_p=1\mathrm{J}{\mathrm{K}}^{-1}$. Suppose together they form an isolated system, with no work or phase transition and negligible contact energy. The first law of thermodynamics and thermal equilibrium giveThe final thermometer reading underestimates the sample's original temperature by about $0.495\mathrm{K}$, although it correctly measures their new common temperature. The disturbance becomes small when the participating probe heat capacity is small compared with the sample's. A large controlled heat bath is a different limit: it supplies or removes heat while approximately maintaining its own temperature.

Waiting time is a separate issue from disturbance. Under a toy Newton's law of cooling model for a uniform probe in a maintained heat bath, a constant contact conductance $G$ givesThe time constant $\tau$ controls this exponential decay. For stipulated values $T_b=350\mathrm{K}$, $T_p(0)=300\mathrm{K}$ and $\tau =5\mathrm{s}$, the probe reads about $331.61\mathrm{K}$ after one time constant and $347.51\mathrm{K}$ after three. These are calculated values, not an instrument specification or measured data. Real probes may have multiple relaxation times and additional heat transfer paths; choose an equilibration criterion from the required accuracy, rather than assuming that a fixed waiting time works universally.

A thermopile connects multiple thermocouples in series to turn a temperature difference into a larger electrical signal. Absorbed thermal radiation can heat its sensing junctions while reference junctions remain cooler; the signal depends on the thermal and optical design.

A thermopile is an array of thermocouples connected as a series circuit with their electromotive forces adding in the chosen terminal order. It measures a collective temperature difference through the Seebeck effect. An open-circuit electric potential difference requires no externally supplied electric current through the array: a high-input-electrical resistance voltmeter reads the signal. The heat that maintains the junction temperature difference still comes from somewhere, such as absorbed electromagnetic radiation.

In a membrane detector, an absorbing film heats the measuring junctions while reference junctions lie on the surrounding support. Actual junction temperatures can differ across the membrane.^[ref] This converts difficult-to-observe heat transfer into an accessible electrical signal; the series connection combines many small contributions without requiring every thermocouple to have a separate meter.

A thermopile absorber converts incident electromagnetic radiation into heat near the thermocouple measuring junctions. Electrically separate junctions can share its temperature through thermal contact. The reference support provides a path for heat transfer and a comparison temperature; “cold” denotes this reference role, even if it is warmer than the absorber.

Distinguish incident power from absorbed power. A window, the illuminated area, and the absorption of the coating determine their relation. A detector's measured voltage sensitivity can depend on where a small illuminated spot lies.^[ref] The model in thermopile thermal response takes absorbed power as its input, so an optical calibration is a separate step.

Use the A-positive terminal order of Seebeck effect for each thermocouple. Its thermocouple measuring junction has temperature $T_{h,i}$ and its reference connections have temperature $T_{c,i}$. Define the difference of Seebeck coefficients as ${\alpha }_i(T)=S_{A,i}(T)-S_{B,i}(T)$. Connect the B reference end of one thermocouple to the A reference end of the next at a common local temperature, so the terminal electric potential differences add:The first A reference terminal is the positive meter input; the final B reference terminal is the negative input. Following this connected path and adding the individual Seebeck effect integrals proves the result. Matched meter leads with common endpoint temperatures cancel as in an isothermal terminal block. The manufacturer's sum formula also allows unequal individual contributions.^[ref]

If all pairs share $T_h,T_c$ and the same nearly constant difference $\alpha$ of Seebeck coefficients throughout that temperature interval, thenThis $\alpha$ is the pair's relative Seebeck coefficient, not either material's individual coefficient or a universal device constant. As an original constructed example, choose $N=20$, $\alpha =40\mu \mathrm{V}{\mathrm{K}}^{-1}$ and $\Delta T=5\mathrm{K}$. Each thermocouple contributes $0.20\mathrm{mV}$ andReversing the meter connections changes the sign. Exchanging $T_h$ and $T_c$ reverses the integrals and hence the electromotive force; this reversible sign response does not make a detector with heat conduction a reversible thermodynamic process. Reversing the wiring of one pair instead gives $(N-2)\alpha \Delta T$, because one contribution is subtracted rather than added. For this example the result is $3.6\mathrm{mV}$, and at equal junction temperatures every contribution vanishes.

The open-circuit condition matters. If the array's total electrical resistance is $R_{\mathrm{pile}}$ and the meter input has electrical resistance $R_m$, the elementary voltage divider model gives $V_m=V_{\mathrm{oc}}{R}_m/(R_m+R_{\mathrm{pile}})$. Thus $R_m\gg R_{\mathrm{pile}}$ is needed to suppress voltmeter loading. Appreciable electric current also permits Joule heating and the thermoelectric heat transfer to disturb the assumed junction temperatures.

Thomas E. Kehl-Fie is a microbiologist whose research concerns metal availability and host defense. His published name is retained without expanding the middle initial. A 2011 collaborative study investigated calprotectin, a host protein involved in nutritional immunity, connecting metal chelation with biological function.^[ref]

The study used isothermal titration calorimetry to measure metal binding to calprotectin. Its thermograms provide evidence about a chemical interaction, complementary to separate functional measurements. This is a useful distinction: binding a nutrient, changing an enzyme's activity and changing a biological outcome are different observations.^[ref]

For an original idealized constant-pressure measurement, suppose one increment yields a measured heat of $-8.0\mu \mathrm{J}$ and a matched analytical blank yields $-2.0\mu \mathrm{J}$. Assume that blank correction removes all nonbinding contributions and that $0.30\mathrm{nmol}$ of additional complexes actually form. The corrected heat isand the corresponding molar enthalpy change isThe sign denotes an exothermic binding contribution under the stated assumptions. These invented numbers are not affinities or measurements from Kehl-Fie's work. A single corrected heat value also does not determine a dissociation constant: that inference requires a binding model and measurements across suitable conditions.

Tidal dissipation converts mechanical energy associated with a time-varying tidal deformation into heat. It can reduce orbital eccentricity, drive synchronous rotation, and contribute deep planetary heating. A circular orbit with exact synchronous rotation, perpendicular spin axis and no additional forcing has a static tidal deformation in the body's frame and need not sustain substantial tidal heating.

Tidal heating converts mechanical energy of tidal deformation into internal heat through tidal dissipation. Sustained orbital eccentricity or a spin axis tilted relative to the orbital plane's perpendicular can maintain changing tidal deformation and heat an exoplanet interior. Thus synchronous rotation alone does not eliminate tidal heating.

Transport phenomena are processes that redistribute quantities such as mass, momentum and energy. Bulk motion produces advection, while microscopic motion and interactions can produce diffusion, heat conduction and viscosity.

Constitutive laws relate a flux to a driving gradient: Fourier's law gives heat flux $\mathbf{q}=-k\nabla T$, and Fick's first law gives diffusive flux $\mathbf{j}=-D\nabla c$. Combining these relations with conservation balances gives evolution equations. The coefficients and the appropriate transported quantity depend on the medium; a formal similarity between two diffusion equations does not make their physical diffusivities identical.

The two-fluid model of superfluidity gives the normal component of a superfluid and the superfluid independent velocities.

An elementary uniform-flow example makes the distinction operational. Write the mass flux per unit area and the ideal entropy flux asHere $s$ is specific entropy, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and ${\rho }_s,{\rho }_n$ are response mass densities. The mass density ${\rho }_s$ used here gives the three-dimensional phase stiffness coefficient ${\rho }_s(\hslash /m{)}^2$ when ${\mathbf{v}}_s=(\hslash /m)\nabla \theta$ is substituted into the kinetic energy density ${\rho }_s{v}_s^2/2$. For a stipulated ${\rho }_s=3\rho /4$, ${\rho }_n=\rho /4$ and $v_n=3u$ along a channel, zero net mass flux requires $v_s=-u$. Yet $j_S=3\rho su$ is positive. Opposing velocities can therefore transport entropy while transporting no net mass.

This example explains why excellent apparent heat transport need not be ordinary diffusion of heat through a stationary liquid. In the ideal zero-net-mass-flux limit, the heat flux associated with this entropy transport is $T{j}_S$. Viscous losses, quantum vortices and external heat exchange require additional terms. The model also supports second sound.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.

For a species $s$ in two well-mixed vessels of equal volume $V$ joined by a thin opening of area $A_h$, the effusion law gives ${\dot{n}}_{s,L}=-(A_h/4V)({\overline{v}}_{s,L}{n}_{s,L}-{\overline{v}}_{s,R}{n}_{s,R})$. Each species' total inventory is conserved. With fixed temperatures and total density sum $n_{s,L}+n_{s,R}=n_{s,\Sigma }$, the left density relaxes exponentially with time constant ${\tau }_s=4V/[A_h({\overline{v}}_{s,L}+{\overline{v}}_{s,R})]$ to $n_{s,L}^{\infty }=n_{s,\Sigma }\sqrt{T_R}/(\sqrt{T_L}+\sqrt{T_R})$.

At equal temperatures, ${\tau }_s=2V/(A_h{\overline{v}}_s)\propto \sqrt{m_s}$, and initially separated species mix at different rates. At unequal temperatures, the final density ratio is $n_{s,L}/n_{s,R}=\sqrt{T_R/T_L}$ for every species, while the partial-pressure ratio is $p_{s,L}/p_{s,R}=\sqrt{T_L/T_R}$. The mass-independent final ratios do not imply mass-independent transient rates. Maintaining different temperatures requires external heat exchange even after net particle transfer vanishes.

Ultraviolet radiation is electromagnetic radiation at shorter wavelengths than visible violet light and longer wavelengths than the neighboring X-ray band. Its photons have higher energies than visible photons because $E=h\nu =hc/\lambda$.

Its interaction with matter depends strongly on wavelength. A material that transmits visible light can absorb ultraviolet radiation, converting excitation into fluorescence, chemical change or heat. To measure transmission, compare the sample with a reference path at the same wavelength and account for the detector's spectral responsivity. Visible transparency is therefore an unreliable test of an ultraviolet window.

For an ideal working gas and $r=V_2/V_1>1$, two isotherms at $T_h,T_c$ produce net work $nR(T_h-T_c)\ln r$. Without a regenerator, isochoric heating requires additional external heat $C_V(T_h-T_c)$, giving the displayed thermal efficiency, strictly below Carnot efficiency. A perfect reversible regenerator recovers this heat and attains the bound instead. Reversing the unregenerated cycle gives cooling coefficient of performance $T_c/(T_h-T_c)-C_V/(nR\ln r)$; positive cooling requires a sufficiently large expansion ratio.

A hypothetical Van der Waals equation model illustrates positive and negative Joule-Thomson coefficient regions and a finite throttle that first warms, then cools. Use the molar convention already distinguished in Van der Waals equation:Here $v_{\mathrm{m}}$ is molar volume, $R$ the molar gas constant, $u_{\mathrm{m}}$ internal energy per mole, and $C_{V,\mathrm{m}}$ the stipulated constant molar heat capacity at fixed volume. The second equation is an additional caloric assumption: an equation of state alone does not specify the temperature dependence of internal energy. It satisfies the equilibrium identity $(\partial u_{\mathrm{m}}/\partial v_{\mathrm{m}}{)}_T=T(\partial p/\partial T{)}_{v_{\mathrm{m}}}-p=a_{\mathrm{m}}/v_{\mathrm{m}}^2$. The arbitrary energy reference is set to zero.

Define dimensionless variables $r=v_{\mathrm{m}}/b_{\mathrm{m}}$, $\theta =R{b}_{\mathrm{m}}T/a_{\mathrm{m}}$, $\pi =b_{\mathrm{m}}^{2}p/a_{\mathrm{m}}$ and $\eta =b_{\mathrm{m}}{h}_{\mathrm{m}}/a_{\mathrm{m}}$, where $h_{\mathrm{m}}$ is molar enthalpy. These differ from the critical-value scaling in Van der Waals law of corresponding states. The throttle calculation reduces toThus a prescribed inlet fixes $\eta$, and solving the first and last equations at the outlet pressure gives the final temperature. For an inlet $({\theta }_1,{\pi }_1)=(0.8,0.12)$ and outlet ${\pi }_2=0.025$, the model gives ${\theta }_2\simeq 0.752287$, a decrease of $0.047713$. The inlet $(2.2,0.12)$ instead gives ${\theta }_2\simeq 2.205786$, an increase of $0.005786$. The purple inlet $(1.45,0.30)$ initially warms but ends at ${\theta }_2\simeq 1.437585$ for the same outlet pressure. These are reduced toy values, not kelvin readings or experimental evidence. No physical gas parameters were fitted.

For a short inversion derivation, implicit differentiation of the Van der Waals equation at fixed pressure givesInsert this into $T(\partial v_{\mathrm{m}}/\partial T{)}_p=v_{\mathrm{m}}$ and substitute back into the equation of state. The Joule-Thomson inversion curve isPositive inversion pressure requires $r>3/2$. Its maximum pressure is $\pi =1/3$ at $r=3$, while the upper dilute limit is ${\theta }_{\mathrm{inv}}\to 2$. Using Van der Waals critical point values, the latter is $T_{\mathrm{inv},\max }/T_c=27/4$. The inversion locus depends on the equation of state; the chosen molar heat capacity additionally determines how much temperature changes along each finite throttle.^[ref]

The displayed region has $\theta >8/27$, above the model's critical temperature, and uses only $r>1$, $p>0$, $(\partial p/\partial v_{\mathrm{m}}{)}_T<0$ and positive $C_{p,\mathrm{m}}$. Subcritical calculations require attention to phase coexistence, metastability and the Maxwell construction; an unstable homogeneous continuation is not an equilibrium throttle endpoint. Even on a stable branch, this mean-field theory with constant $C_{V,\mathrm{m}}$ is a qualitative model, not an accurate inversion map for a particular real gas. Phase transitions, changing composition, appreciable heat leakage or bulk kinetic energy changes require a different property calculation or steady-flow energy balance. At phase coexistence, the smooth single-thermodynamic phase coefficient formula cannot replace a balance involving the phase fraction.

For an ideal Diesel cycle with compression ratio $r=V_1/V_2>1$, cutoff ratio $\rho =V_3/V_2>1$ and constant heat-capacity ratio $\gamma$, its thermal efficiency is $\eta =1-({\rho }^{\gamma }-1)/[\gamma r^{\gamma -1}(\rho -1)]$. As $\rho \to 1^{+}$ the quotient tends to $\gamma$, giving $\eta \to 1-r^{1-\gamma }$. The maximum-to-minimum temperature ratio tends to $r^{\gamma -1}$, so the limiting efficiency equals that of a Carnot cycle between those temperatures. Both heat input and net work vanish in this limit: the result is a limit of their ratio, not a finite-work engine at exactly zero cutoff interval.

Vapor-compression refrigeration circulates a working fluid through cold and hot coils. In the cold coil, liquid becomes gas while absorbing heat from the refrigerated region. Compression supplies work and raises the pressure and temperature; the hot coil releases heat as the gas returns to liquid. A restriction reduces the pressure before the next cold-side passage. Changing the pressure changes the temperature of the phase transition, letting one circulating fluid receive and release heat at different temperatures. This physical implementation serves both refrigerators and heat pumps; its ordinary dissipative operation differs from the reversible Carnot cycle used to bound performance. ^{[ref] [ref]}

For an ideal two-layer natural ventilation transient following a heating change $Q\mapsto \lambda Q$, let $V_0$ and ${\theta }_0$ be the initial flow and temperature excess, $s=V_{0}t/(A_{r}H)$, $\eta =h/H$, and $x=\eta (1-\theta /{\theta }_0)$. The integrated buoyancy pressure gives $V/V_0=\sqrt{1-x}$, while heat balance gives $x_s=\sqrt{1-x}-\lambda$. A material interface satisfies ${\eta }_s=\sqrt{1-x}$, so $\eta =x+\lambda s$. For $0<\lambda <1$, $x$ increases toward $1-{\lambda }^2$ and ${\eta }_s\ge \lambda$, ensuring finite-time filling of the room. After filling, $y=\theta /{\theta }_0$ obeys $y_s=\lambda -y^{3/2}$ and tends to ${\lambda }^{2/3}$. This model assumes well-mixed individual layers, negligible wall heat capacity and a sharp advected interface.