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Cold-junction thermal coupling is the heat-transfer relationship between a thermocouple reference junction, its temperature sensor and the surrounding apparatus. Cold-junction compensation needs the actual terminal-region temperature. A sensor with excellent electrical calibration can still give an unsuitable correction if it is at another temperature because of heat conduction, air currents, nearby dissipation or finite response time.

An isothermal terminal block aims to keep both thermoelement-to-lead transitions and the reference sensor at a common temperature while preserving electrical insulation. The TI design discussion identifies two separate requirements: matching the temperatures of the two terminal transitions and measuring that temperature accurately. Its layout discussion considers metal fill, parasitic heating and airflow.^[ref] NI's documented terminal assemblies likewise incorporate local temperature sensors rather than inferring the terminal temperature from the raw thermocouple signal.^[ref]

A simple lumped thermal model separates the paths. Let $T_j$ be a prescribed junction temperature, $T_s$ the sensor temperature, $T_a$ ambient temperature, $G_c$ the thermal conductance to the junction, $G_a$ the conductance to ambient and $P_s$ sensor dissipation. With sensor heat capacity $C_s$,For fixed inputs, the steady sensor error isBetter coupling to the junction and weaker parasitic coupling to ambient have different roles; sensor dissipation adds a further term.

For an original example choose $T_j=35{}^{\circ }\mathrm{C}$, $T_a=25{}^{\circ }\mathrm{C}$, $G_c=0.040W/K$, $G_a=0.010W/K$ and $P_s=1\mathrm{mW}$. ThenThe sensor settles near $33.02{}^{\circ }\mathrm{C}$ despite the junction being at $35{}^{\circ }\mathrm{C}$. These are stipulated thermal parameters, not specifications of the cited devices.

The same equation gives thermal time constant $\tau =C_s/(G_c+G_a)$. With $C_s=0.10J/K$, $\tau =2\mathrm{s}$. Waiting several time constants removes the transient toward the steady sensor temperature, but cannot remove the calculated steady mismatch. A stable display is therefore not proof of thermal agreement.

The reference-junction temperature sweep in thermocouple testing can expose both effects. At matched plateaus, compare the reference reading with a local independent thermometer; during ramps, compare lag and sweep-direction dependence. If the two reference terminals themselves have different temperatures, even one perfectly measured sensor temperature cannot satisfy the simple two-junction compensation model. The thermal configuration must be tested as part of the measurement chain.

Take a stipulated pair of effective rods with common length $L$, linear thermal expansion coefficient ${\alpha }_L$, heat capacity $C$, projected illuminated area $A$, and thermal conductance $G$ to an environment at constant temperature $T_a$. One reference represents the symmetric polished-rod assembly. Write ${\theta }_b=T_b-T_a$ and ${\theta }_r=T_r-T_a$. Let their effective radiometric absorptances be $a_b>a_r$ for a fixed incident optical spectrum, and let $E(t)$ be the common irradiance. A spatially uniform, linear heat-balance model givesHere $G$ includes a linear approximation to the relevant losses; changing the surface treatment can also change those losses in a real device. We stipulate equal $C$ and $G$ to isolate the response to absorbed illumination, with thermal coupling from the burner neglected. Small, mechanically unconstrained extensions then give a differential displacementThe time constant $\tau$ describes thermal memory. A common imposed temperature shift cancels from the extension difference under the matched-coefficient assumptions; mismatched materials, geometry or thermal coupling can spoil that cancellation.

For a step from darkness to constant irradiance $E_0$, with initially equal rod temperatures, the first-order linear differential equation givesChoose $L=0.20\mathrm{m}$, ${\alpha }_L=20\times 10^{-6}{\mathrm{K}}^{-1}$, $A=0.0020{\mathrm{m}}^2$, $a_b=0.9$, $a_r=0.1$, $C=5.0\mathrm{J}{\mathrm{K}}^{-1}$, $G=0.20\mathrm{W}{\mathrm{K}}^{-1}$ and $E_0=600\mathrm{W}{\mathrm{m}}^{-2}$. The calculated difference of absorbed powers is $0.96\mathrm{W}$, givingFor an illustrative ideal valve that closes at $\delta =g=10\mu \mathrm{m}$, the closing time isThis threshold is crossed only when $0<g<{\delta }_{\infty }$. These are chosen model parameters, not a fit to a historical valve.

After illumination stops at $t=t_1$, the displacement decays as $\delta (t)=\delta (t_1)e^{-(t-t_1)/\tau }$. With $t_1=180\mathrm{s}$, the model reopens about $16.3\mathrm{s}$ later. Thus a sun valve responds to the thermal history of illumination; it does not read a clock or directly measure the beacon's brightness.

A lumped sample of heat capacity $C$ coupled by thermal conductance $G$ to a bath has the displayed linearized energy balance. For constant internal power, the bath-to-sample temperature response is $1/(1+i\omega C/G)$. Insulation and heat capacity suppress rapid bath fluctuations but have unity static bath-to-sample gain. A steady power change shifts the sample by $\delta P/G$, so improved isolation increases self-heating sensitivity and equilibration time. The lumped approximation requires internal equilibration faster than the variations being modeled.

The equilibrium calculation determines where the two temperatures end, without determining how long they take. Stipulate a constant thermal conductance $G>0$ and the linear heat transfer law underlying Newton's law of cooling. With $Q$ positive from body 1 to bath 2, the ordinary differential equations areThe dot denotes a derivative with respect to time. This model assumes internal equilibration is fast compared with exchange between the bodies. Adding the two energy rates gives zero; subtracting the temperature rates givesThe temperature difference has exponential decay with time constant $\tau$. For the finite heat bath example and the stipulated $G=2\mathrm{W}{\mathrm{K}}^{-1}=2\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{K}}^{-1}$, $\tau =45\mathrm{s}$, and solving the ordinary differential equations givesAt one time constant, $T_1\simeq 315.865\mathrm{K}$, $T_2\simeq 293.793\mathrm{K}$ and $Q\simeq 3413.451\mathrm{J}$. Equal temperatures are approached as a limit at infinite time. The finite bath causes the temperature difference to shrink faster than the fixed-bath model, because both bodies change temperature.

For an ideal heat bath with the same $G$, set $T_2=T_b=290\mathrm{K}$ rather than evolving its temperature. Newton's law of cooling then yieldsChanging $G$ changes these time constants, while the final temperature is fixed by the capacities and initial conditions. The dynamical model independently checks the Second law of thermodynamics:Its integral over the relaxation reproduces the endpoint entropy change. The useful separation is between capacity, which governs how much the bath temperature shifts, and thermal conductance, which governs the exchange rate. Neither alone establishes a reliable heat bath.

A controlled thermal link: high thermal conductance lets a cold stage equilibrate with a heat bath; low thermal conductance suppresses heat transfer during adiabatic demagnetization.

Three distinct effects spoil indefinite $T\propto B$ cooling: extra entropy outside the spin subsystem, additional magnetic energy scales, and irreversible heat or work input. They can be distinguished with a thermodynamic derivative. For a smooth equilibrium Helmholtz free energy obeying $dF=-SdT-mdB$, a Maxwell relation givesHere $C_B$ is the total heat capacity participating in the stage's thermal equilibrium. The ideal model has $(\partial m/\partial T{)}_B=-C_{B,\mathrm{mag}}/B$. If its spins share temperature with a field-independent lattice and load of combined heat capacity $C_{\mathrm{other}}>0$, thenThe positive slope still means cooling when $dB<0$, but its smaller magnitude gives less cooling per fractional field reduction. For example, a low-temperature Debye model gives a lattice entropy proportional to $T^3$. As that lattice entropy falls, the spins take up the released entropy; $x$ can no longer remain constant. A weak enough coupling to neglect its equilibrium interaction energy can still be strong enough to exchange heat on the sweep timescale.

Interactions between magnetic moments, including exchange interaction and the fields of neighboring magnetic dipole moments, introduce energy scales that do not vanish with the applied $B$. Once they compete with $k_{B}T$ or $\mu B$, the independent-site partition function fails. Correlations and possible magnetic ordering alter the available magnetic entropy and the cooling curve. Simply setting the external magnetic field to zero does not eliminate those interactions. Different refrigerants therefore need their own measured equilibrium properties; there is no universal final temperature from the applied-field ratio alone.

Finally, a real open heat switch has residual thermal conductance; supports, wires and radiation carry incoming heat. Field changes may also dissipate energy. Locally near thermal equilibrium, with incoming heat rate ${\dot{Q}}_{\mathrm{leak}}$ and internal entropy production rate ${\dot{S}}_{\mathrm{prod}}$, the stage balance becomesThis uses a common stage temperature; a non-equilibrated load requires separate balances. The sweep must allow spin-lattice relaxation while remaining fast enough to limit accumulated parasitic heat. At fixed $B$, the same balance explains warming and finite hold time. Useful refrigeration is the controlled absorption of a load's heat during that finite interval, followed by a renewed isothermal process of magnetization to export its entropy.

Low-temperature calorimetry measures a small sample's heat capacity by resolving a small temperature change after a known electrical energy input. The experimental challenge is that the support, heater and thermometer also warm up. Their calorimeter addenda must be removed before interpreting the sample's phonons or electronic heat capacity. The Debye T-cubed test turns the corrected response into a test of the low-temperature vibrational spectrum.

Manuel Brando's 2009 experiment used a sapphire platform suspended by four alloyed platinum wires from a copper ring fixed to a cryostat stage. Grease attached the sample; a chip heater and ruthenium-oxide resistance thermometer served platform PLS-2. A controller stabilized the bath; a lock-in amplifier supplied heating pulses and read the calibrated sensor. The steady rise calibrated thermal conductance as $G=P/\Delta T$. A separate $59.1\mathrm{mg}$, $99.99\%$ pure gold sample, compared with reference heat capacities, established the calorimeter addenda. Small pulses limited property variation; imperfect sample contact required a second relaxation time. The platform contribution rose below $1.5\mathrm{K}$, making simple extrapolation unsafe.^[ref]

For an original idealization, assume sample and platform share one temperature $T$, the bath remains at $T_b$, and both the total heat capacity and the weak-link thermal conductance $G$ are approximately constant across one small pulse. Ignore other heat transfer and any phase transition. The first law of thermodynamics becomesHere $C_s$ is sample heat capacity, $C_a$ is calorimeter addenda, $P$ is deposited electrical power from Joule heating, and $\tau$ is a time constant. This is the same balance as calorimeter heat-loss correction, used here with a deliberately weak cryogenic link and a separately subtracted platform. For a constant pulse $P_0$ beginning at bath temperature, then switched off at $t_h$,The cooling exponential decay determines $\tau$, while the heating plateau determines $G=P_0/(T_{\infty }-T_b)$. Together with calorimeter addenda, they give $C_s=G\tau -C_a$. Cooling alone determines a ratio, not an absolute heat capacity. With finite $t_h$, the observed end rise is smaller than the plateau; fit the full response or apply the displayed correction rather than identifying the two.

For the stipulated figure parameters, $G=5.00\times 10^{-8}\mathrm{W}{\mathrm{K}}^{-1}$, $\tau =20.0\mathrm{s}$, and $C_a=2.00\times 10^{-7}\mathrm{J}{\mathrm{K}}^{-1}$ give $C_s=8.00\times 10^{-7}\mathrm{J}{\mathrm{K}}^{-1}$. The asymptotic rise is $0.0400\mathrm{K}$, only $2\%$ of bath temperature. At $100\mathrm{s}$, the rise is $0.0397305\mathrm{K}$. Of $2.00\times 10^{-7}\mathrm{J}$ input, only $3.97305\times 10^{-8}\mathrm{J}$ remains stored: the other $1.60270\times 10^{-7}\mathrm{J}$ has flowed to the bath. Treating this long pulse as an isolated $Q/C$ experiment would overestimate total heat capacity by a factor of $5.034$.

Suppose independent standard uncertainties are $u(G)=0.05\times 10^{-8}\mathrm{W}{\mathrm{K}}^{-1}$, $u(\tau )=0.2\mathrm{s}$ and $u(C_a)=0.10\times 10^{-7}\mathrm{J}{\mathrm{K}}^{-1}$. First-order propagation of uncertainty givesThe uncertainty is one standard deviation under those assumptions, about $2.2\%$ of sample heat capacity. Shared thermometer calibration errors can correlate fitted $G$ and $\tau$; then their covariance contributes too. Smaller samples amplify the relative influence of calorimeter addenda. Proposed checks include varying pulse amplitude, repeating with a reference sample, checking sensor self-heating, and comparing single- and two-time-constant fits. A drifting bath, slow internal equilibration or a transition invalidates this simple reduction even if a cooling curve looks smooth.

A lumped thermal model assigns one representative temperature and heat capacity to each region of an apparatus. For a single body of heat capacity $C$ connected to a bath by thermal conductance $G$, with supplied power $P$, an elementary balance isThe model stores internal energy in the body and treats exchange through a thermal resistance, leaving internal spatial temperature gradients unresolved.

A small solid specimen with rapid internal equilibration can approximate this behavior. A constant heating step gives a final rise $P/G$ and a thermal time constant $C/G$ when $C$ and $G$ are constant. Comparing thermometers at different positions tests the assumption of nearly uniform specimen temperature. If appreciable internal gradients persist, the apparatus needs several coupled thermal regions or a spatial heat equation model. Fitting a single exponential is useful only after this physical approximation has been checked.

A microbolometer is a miniaturized bolometer in which absorbed electromagnetic radiation heats a thermally isolated sensing element and the readout detects the resulting change in electrical resistance. Arrays of microbolometers can form an infrared image without cryogenic detector cooling. A pixel commonly has an absorbing membrane suspended above its readout circuit on narrow supports.^[ref]

An elementary small-signal heat transfer model writesHere $C$ is the sensing element's heat capacity, $G$ its thermal conductance to a fixed-temperature substrate $T_b$, $P$ incident radiant power and $\eta$ its absorbed fraction. The model treats the element as having a single temperature and neglects changes in readout self-heating. A step to constant incident power $P_0$ givesThe thermal time constant therefore connects sensitivity to response speed. Lowering $G$ increases the steady temperature rise, but at fixed $C$ also lengthens the response.^[ref]

For an original example, choose $C=10^{-9}\mathrm{J}{\mathrm{K}}^{-1}$, $G=10^{-7}\mathrm{W}{\mathrm{K}}^{-1}$, $\eta =0.8$ and $P_0=1\mathrm{nW}$. Then $\tau =10\mathrm{ms}$ and the final sensing-element rise is $8\mathrm{mK}$; after $30\mathrm{ms}$ it has reached about $95\%$ of that rise. If the initial resistance is $100\mathrm{k\Omega }$ and the temperature coefficient of resistance is $\alpha =-0.02{\mathrm{K}}^{-1}$, the linear approximation $\Delta R\approx \alpha R\theta$ gives $\Delta R=-16\Omega$ at steady state. These stipulated values illustrate the model, not a particular camera's specification.

The sensing element's temperature rise is not the observed object's temperature rise. Inferring object temperature from the measured infrared signal additionally depends on emissivity, reflected radiation, the optical path and measurement calibration. “Uncooled” also permits temperature stabilization of the array; it describes the absence of required cryogenic operation, not the absence of thermal design.

Use an enclosure with thermal insulation and sufficient heat capacity to slow the response to ambient changes. Block draughts and direct radiant heating, use low-conductance supports and thermal breaks in wiring, and reduce thermal radiation with low-emissivity shields. Vacuum suppresses gas thermal conduction and convection but leaves thermal radiation and solid thermal conduction. A conducting inner shield helps make temperature uniform; it is not by itself thermal insulation.

For a lumped load of heat capacity $C_{\mathrm{th}}$ coupled to the room by thermal conductance $G$, the first-order thermal isolation model isLarge thermal resistance and heat capacity attenuate fast fluctuations; nested shields can provide several filtering stages. Passive thermal insulation does not reject arbitrarily slow ambient drift, since this transfer approaches one at zero frequency. It also increases the temperature rise $\delta P/G$ caused by internal dissipation.

For long-term stability use a regulated bath or thermostat, placing a calibrated sensor near the sensitive component. Heater feedback rejects slow disturbances, provided sensor lag, loop stability, actuator range and electrical pickup are controlled. Maintain stable internal power, allow equilibration, record both room and load temperatures, and test rejection by a deliberate ambient-temperature step or sinusoidal modulation. These measurements reveal gradients and thermal bypasses that an thermal insulation label alone cannot establish.

temperature changes alter dimensions, resistances, refractive indices and offsets; gradients can generate thermoelectric voltages. Identify the dominant heat paths and the experiment's temperature coefficients before choosing insulation. Reduce convection with an enclosure or, where appropriate, vacuum; reduce conduction through supports and leads with suitable geometry and thermal anchoring; use thermal radiation shields and low-emissivity surfaces to reduce radiative exchange. vacuum alone does not remove heat conduction through wires or thermal radiation.

In first-order thermal isolation, a sample with heat capacity $C$ and thermal conductance $G$ to its surroundings obeysThermal mass and insulation suppress rapid fluctuations but only delay slow ambient drift: the zero-frequency gain remains one. Active temperature control, a stable reference environment, adequate equilibration time and an appropriate sensor location address long-term stability. Self-heating produces a steady offset $\delta T_s=\delta P/G$; increasing insulation can worsen that offset and settling time. Check temperature gradients rather than relying on a remote thermometer.

Match materials or use differential/reference measurements to reduce expansion and resistance drift; keep dissimilar-metal junctions isothermal to suppress thermoelectric offsets. Log temperature alongside the signal and deliberately vary environmental temperature to measure residual coupling, using a physically justified correction with uncertainty. Interleave calibrations and allow equilibration after changing power. Feedback must remain stable, and heater/sensor electronics should not introduce electrical interference into the measurement.

Temperature changes affect measurements through several routes. Thermal expansion changes lengths and alignments; temperature-dependent resistance and refractive index change electrical and optical calibrations; temperature differences between dissimilar junctions generate Seebeck effect voltages. These slow environmental changes are distinct from unavoidable equilibrium thermal fluctuations and Johnson-Nyquist noise. A useful strategy identifies the coupling to the measured signal before choosing a remedy.

Reduce environmental coupling by enclosing the apparatus against drafts, using thermal insulation and radiation shields, isolating heat sources, and minimizing dissipated power. A vacuum enclosure suppresses convection but leaves thermal conduction and thermal radiation paths to control. A large heat capacity reduces rapid temperature excursions. In a simple thermal model with heat capacity $C_T$ and thermal conductance $G_T$ to a reservoir, the relaxation time is $C_T/G_T$, so adequate settling time is necessary after changing the apparatus. A temperature sensor close to the sensitive element and stable negative feedback heating or cooling can regulate its temperature; poorly placed sensors or excessive feedback gain can leave gradients or cause oscillations.

Electrical thermal noise has the classical low-frequency one-sided voltage power spectral densityfor a resistor measured over a flat bandwidth $B$ without significant loading. More generally replace $B$ by the effective noise bandwidth. Lower temperature or narrower bandwidth reduces the noise, and resistance choices must account for source loading and the amplifier's own voltage and current noise. A lock-in amplifier uses modulation and narrowband detection to reject much low-frequency drift and out-of-band noise, while thermal noise inside the accepted band remains. Temperature stabilization does not remove equilibrium fluctuations.

Use a matched reference or differential bridge to reject common environmental changes. Keep junctions isothermal and choose matched materials to reduce thermoelectric offsets. For a resistance measurement, reversing a small test current separates a stable offset from the wanted voltage:Reversal must be fast enough that the offset stays approximately fixed, with enough settling time and sufficiently little self-heating. Recording the sample temperature permits a measured calibration slope $dS/dT$ to correct a signal to a reference temperature; the correction's calibration and thermometer uncertainties must be propagated. Independent white noise fluctuations decrease under averaging, whereas correlated temperature drift may not decrease as the inverse square root of the number of readings. Temperature records and repeated reference measurements help distinguish the two.

Space heating supplies energy to an occupied region to maintain or raise its temperature. The thermal balance includes heat transfer through walls, exchange with ventilating air, stored internal energy and internal sources. A simple lumped thermal model writes $C\dot{T}=P-G(T-T_{\mathrm{out}})$ when all losses can be approximated by one effective thermal conductance.

At steady temperature, supplied power balances the ongoing losses; during warm-up, some also increases stored internal energy. Hot-water heating, steam heating and hot-air heating differ in how energy is carried to the room. A heat pump can deliver heat partly extracted from outside, whereas an electrical resistance heater converts supplied electrical work locally. Comparing room temperature with input power after a small controlled step tests both the effective loss coefficient and the thermal time constant of the simplified model.

A temperature sensor is a sensor whose physical response is used to estimate temperature. A resistance thermometer and a thermistor use changes in electrical resistance; a thermocouple uses the Seebeck effect. The sensor element, its thermal contact and its electrical readout together determine what a temperature measurement means.^[ref]

A contact sensor first responds to its own temperature, which need not immediately equal that of the object or fluid. A useful original lumped thermal model writes the first law of thermodynamics for a sensor with heat capacity $C$, thermal conductance $G$ to an environment at $T_e$, and internal heating power $P$:For constant inputs, its thermal time constant and equilibrium are $\tau =C/G$ and $T_{s,\infty }=T_e+P/G$. The temperature sensor response time and the bias from heating are different properties. Better thermal coupling both shortens $\tau$ and reduces $P/G$ in this model.

Take stipulated values $C=0.020J/K$ and $G=0.010W/K$. Then $\tau =2\mathrm{s}$. With $P=0$ and an environment step from $20$ to $30{}^{\circ }\mathrm{C}$,At $t=2\mathrm{s}$ the sensor reads about $26.32{}^{\circ }\mathrm{C}$, despite a perfect electrical readout. Reaching $90\%$ of the step takes $\tau \ln 10=4.605\mathrm{s}$. A steady $1\mathrm{mW}$ heating input would instead add $0.10\mathrm{K}$ at equilibrium. These are model calculations, not universal sensor specifications.

A proposed two-bath step test of a temperature sensor transfers a small probe between two well-stirred baths with independently established temperatures and records its readout against time. Repeating with a protective sleeve tests the thermal part of the measurement chain. The DOE handbook illustrates insulated sensor elements inside metal sheaths and protective wells; electrical accuracy alone does not describe that complete assembly.^[ref]

Thermal loading by a contact temperature sensor is another limit: a probe exchanges heat with the object and can perturb the quantity it is meant to measure. A small object may require a coupled thermal model rather than a fixed $T_e$. For a resistance thermometer, the excitation also causes Joule heating; checking several currents can distinguish that effect from a fixed readout offset.^[ref]

A temperature sensor response time specifies how long a temperature sensor takes to complete a stated fraction of its response to a change in the surrounding temperature. The fraction and the test conditions are part of the specification: a 90% response time and a 63.2% response time describe different points on the same record. Rapid immersion in a different-temperature liquid approximates the required step input.^[ref]

For a lumped thermal model with sensor heat capacity $C$, constant thermal conductance $G$ to a bath at $T_b$, and negligible internally generated heat, the energy conservation equation and its solution areThus the time constant $\tau$ is the time to complete $1-e^{-1}\approx 63.2\%$ of the change. The time to complete a fraction $f$ is $t_f=-\tau \ln (1-f)$, so $t_{90}=\tau \ln 10$, whereas the elapsed time between 10% and 90% is $\tau \ln 9$. For example, $\tau =3\mathrm{s}$ gives $t_{90}\approx 6.91\mathrm{s}$ and $t_{99}\approx 13.82\mathrm{s}$.

A two-bath step test of a temperature sensor tests this exponential function model rather than assuming it. Several internal heat capacity elements, a protective sheath, or electronic filtering can produce several response stages. The thermometry guide reports fast and slow components in tested platinum thermometer assemblies; a single quoted time constant need not describe their entire approach to equilibrium.^[ref]

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.

thermal insulation reduces unwanted heat exchange. Conduction through supports, convection in gas and thermal radiation require different treatments. vacuum suppresses gas transport but leaves solid conduction and radiation; lowering thermal conductance also increases the temperature rise caused by a fixed internal heat load.

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]

A thermal time constant characterizes exponential relaxation in a first-order lumped thermal model. For heat capacity $C$ and thermal conductance $G$ to a fixed bath, a small temperature deviation obeys $C\dot{\delta T}=-G\delta T$, givingAfter one $\tau$, the initial deviation has fallen to $1/e$ of its original value. The ratio has units of time because $C$ is measured in $\mathrm{J}{\mathrm{K}}^{-1}$ and $G$ in $\mathrm{W}{\mathrm{K}}^{-1}$.

A heating step provides an experimental check: the body approaches its final temperature rise with the same $\tau$ in this linear model. Greater heat capacity slows changes, while greater thermal conductance speeds exchange. Several coupled regions generally produce several relaxation modes, so fitting one exponential over a short interval does not establish a unique time constant for the entire apparatus.

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]