Thermal equilibrium

A thermostat senses temperature and adjusts a heating or cooling actuator to approach a reference temperature. For precise regulation, sensor placement, thermal delay and feedback stability matter as well as the nominal reference. Thermal shielding can reduce rapid environmental disturbances while feedback corrects slower ones.

Thermal equilibration of two finite bodies ( $T_{f} = (C_{1} T_{1, 0} + C_{2} T_{2, 0}) / (C_{1} + C_{2})$ )

Consider two rigid bodies forming an isolated system, with fixed composition and constant whole-body heat capacity at constant volume values

C_{1} = 1 kJ K^{- 1}

and

C_{2} = 3 kJ K^{- 1}

. Their initial temperatures are

T_{1, 0} = 300 K

and

T_{2, 0} = 400 K

. Assume each body remains internally uniform, their contact stores negligible energy, no phase transition occurs and no work is done or extracted. These are stipulated model values, not material measurements.

Because

d U_{i} = C_{i} d T_{i}

, write

U_{i} = a_{i} + C_{i} T_{i}

with arbitrary fixed reference constants

a_{i}

. The conserved relative internal energy is

E = U_{1} + U_{2} - a_{1} - a_{2} = C_{1} T_{1} + C_{2} T_{2} = 1500 kJ .

This is an energy reference convention, not a claim about actual heat capacities down to zero temperature. At thermal equilibrium the two temperatures equal

T_{f}

, so the first law of thermodynamics gives

T_{f} = \frac{C _{1} T _{1, 0} + C _{2} T _{2, 0}}{C _{1} + C _{2}} = 375 K, Δ U_{1} = 75 kJ, Δ U_{2} = - 75 kJ .

The larger heat capacity pulls the common temperature closer to its initial value: body 1 warms by

75 K

, while body 2 cools by

25 K

Figure 1.
Two finite bodies approach thermal equilibrium at $375 K$
. Ideal model with $C_{1} = 1$ and $C_{2} = 3 kJ K^{- 1}$ and contact conductance $G = 0.1 kJ s^{- 1} K^{- 1}$ . The weighted temperature stays at $375 K$ and the relative internal energy stays at $1500 kJ$ ; the plotted curves are model predictions. **Details**.

To predict a timescale, stipulate a constant contact conductance

G = 1/ R_{th} = 0.1 kJ s^{- 1} K^{- 1}

, the reciprocal of the contact's thermal resistance. The rate of heat transfer into body 1 is

G (T_{2} - T_{1})

, giving

C_{1} \dot{T}_{1} = G (T_{2} - T_{1}), C_{2} \dot{T}_{2} = - G (T_{2} - T_{1}) .

Adding proves conservation of

E

. Subtracting gives exponential decay of the temperature difference, with time constant

τ = \frac{C _{1} C _{2}}{G ( C _{1} + C _{2} )} = 7.5 s, T_{2} - T_{1} = 100 K e^{- t / τ},

and therefore

T_{1} (t) = 375 K - 75 K e^{- t / τ}, T_{2} (t) = 375 K + 25 K e^{- t / τ} .

Exact equality is approached as

t \to \infty

in this ideal model. Changing

G

changes the rate but not

T_{f}

: equilibrium and the speed of reaching it are separate predictions. Real bodies can have internal temperature gradients, changing heat capacities and contact losses, so the plotted

7.5 s

is an assumed model timescale.

Tolman–Ehrenfest relation ( $T - k^{2} = constant$ )

In static thermal equilibrium, the local temperature times the stationary lapse is constant. Energy measured locally redshifts to conserved Killing energy by that same lapse; requiring a single equilibrium Boltzmann exponent gives the relation. The Killing vector field normalization fixes the constant but not the ratio of temperatures at different locations.

Boltzmann suppression ( $e^{- m / T}$ )

Boltzmann suppression is the exponential reduction of an equilibrium population when creating it costs energy much greater than the temperature. For a nonrelativistic species with negligible chemical potential, the leading factor is

e^{- m / T}

, multiplied by its momentum phase-space volume. It concerns the equilibrium abundance; freeze-out can preserve a larger population than the instantaneous equilibrium value at later times.

Boltzmann suppression of a nonrelativistic relic ( $n_{eq} ≃ g (m T /2 π)^{3/2} e^{- m / T}$ )

A nonrelativistic species with

g

states and negligible chemical potential has equilibrium density

g (m T / (2 π))^{3/2} e^{- m / T}

. If particle-antiparticle annihilation maintains equilibrium after the species becomes massive, its eventual freeze-out population can be much smaller than a relativistically decoupled population. The nonrelativistic annihilation rate and any asymmetry determine the final abundance.

Kinetic equilibrium

Momentum-exchange collisions maintain a common temperature and an equilibrium-shaped momentum probability distribution. Particle abundances and chemical potentials need not have their chemical equilibrium values: elastic scattering alone conserves particle number. Thus kinetic equilibrium does not by itself imply either zero chemical potential or conserved comoving entropy.

Heat bath

A heat bath, or thermal reservoir, is an idealized system so large that exchanging a finite amount of heat does not appreciably change its temperature.

The ideal heat bath supplies a fixed absolute temperature

T_{b}

while its internal energy changes through heat transfer. A real body's suitability as a heat bath depends on its heat capacity relative to the exchange being studied, and on how quickly its internal temperature gradients disappear. Large physical size alone supplies neither condition. A finite heat bath makes the approximation quantitative; a regulated calibration bath instead uses feedback control to maintain its temperature by exchanging energy with external equipment.

For a weakly coupled system that exchanges energy but no matter with an ideal heat bath, the canonical ensemble assigns the Boltzmann distribution

p_{n} = \frac{e ^{- E_{n} / (k_{B} T_{b})}}{Z}, Z = \sum_{m} e^{- E_{m} / (k_{B} T_{b})} .

Here

E_{n}

is a microstate's energy,

k_{B}

the Boltzmann constant and

Z

the partition function. The fixed temperature removes the need to track every exchange with the environment. A substantially changing bath temperature no longer gives this one fixed Boltzmann distribution.^[ref]

Finite heat bath ( $Δ T_{b} = Q_{b} / C_{b}$ )

A finite heat bath has a finite whole-body heat capacity

C_{b}

and therefore changes temperature when it absorbs heat

Q_{b}

. For a rigid closed system with no work, fixed composition and no phase transition, the first law of thermodynamics gives

Q_{b} = Δ U_{b} = \int_{T_{b, 0}}^{T_{b, f}} C_{b} (T) d T .

If the heat capacity at constant volume

C_{b}

is constant throughout this temperature interval, the integral becomes

Q_{b} = C_{b} (T_{b, f} - T_{b, 0})

. This statement concerns an internal energy difference;

C_{b} T_{b}

need not be the body's absolute internal energy.

For an original example, put a rigid body 1 and a finite bath 2 into thermal contact inside an isolated system. Stipulate constant whole-body heat capacities

C_{1} = 100 J K^{- 1}

and

C_{2} = 900 J K^{- 1}

, with initial temperatures

T_{1, 0} = 350 K

and

T_{2, 0} = 290 K

. Each body remains internally uniform; the connection stores negligible energy; no external work, matter exchange or phase transition occurs. These are constructed model values. The argument in thermal equilibration of two finite bodies applies directly: equal final temperatures and conserved total internal energy give

C_{1} (T_{f} - T_{1, 0}) + C_{2} (T_{f} - T_{2, 0}) = 0, T_{f} = \frac{C _{1} T _{1, 0} + C _{2} T _{2, 0}}{C _{1} + C _{2}} = 296 K .

Thus the heat

Q

transferred from body 1 to bath 2 is

Q = C_{1} (350 - 296) K = C_{2} (296 - 290) K = 5400 J .

Body 1 loses

5400 J

of internal energy and bath 2 gains exactly that amount. Body 1 cools by

54 K

, while bath 2 warms by

6 K

. Equal heat transfers produce unequal temperature changes because the heat capacities differ.

Figure 1.
Finite bath temperature changes and the integrated heat ledger
. Original constant-heat capacity model, not measured data. Left: accumulated heat $Q$ leaving body 1 lowers its temperature and raises bath 2's temperature; the lines meet at $Q = 5400 J$ and $T_{f} = 296 K$ . Right: the shaded integrals $\int C_{i} d T$ have equal area $5400 J$ , although their widths are $54 K$ and $6 K$ . The outgoing heat magnitude is shown positive; body 1's signed internal energy change is negative. **Details**.

The heat bath approximation can now be assessed against a chosen allowed temperature change

δ T > 0

. Constant

C_{b}

requires

∣ Q_{b} ∣/ C_{b} \leq δ T

, rather than merely calling the bath large. For this example, an allowed change of

0.1 K

would fail because the computed change is

6 K

. The exact bath shift and its ideal-reservoir limit are

T_{f} - T_{2, 0} = \frac{C _{1}}{C _{1} + C _{2}} (T_{1, 0} - T_{2, 0}), lim_{C_{2} \to \infty} T_{f} = T_{2, 0} = 290 K .

Hold

C_{1}

and both initial temperatures fixed in this limit. The transferred heat then approaches

C_{1} (T_{1, 0} - T_{2, 0}) = 6000 J

. The finite bath receives less heat because its warming reduces the available temperature difference. For heat capacities that vary, use the two integrals in the energy balance instead of this weighted average.

Conservation of energy also permits imagining the reverse transfer, so it cannot explain the direction of equilibration by itself. Entropy from heat capacity gives the additional test, using reversible comparison paths between equilibrium endpoints:

Δ S_{1} = C_{1} ln \frac{296}{350} ≃ - 16.75737 J K^{- 1}, Δ S_{2} = C_{2} ln \frac{296}{290} ≃ 18.43068 J K^{- 1},

Δ S_{tot} = C_{1} ln \frac{T _{f}}{T _{1, 0}} + C_{2} ln \frac{T _{f}}{T _{2, 0}} ≃ 1.673308 J K^{- 1} > 0.

(6)

All ratios inside the natural logarithms use absolute temperatures. The actual exchange across unequal temperatures is irreversible, even though reversible paths calculate the endpoint entropy differences.

There is a short general proof of positivity. Put

w_{i} = C_{i} / (C_{1} + C_{2})

for positive constant heat capacities and positive initial temperatures. The strict concavity of the natural logarithm gives

\frac{Δ S _{tot}}{C _{1} + C _{2}} = ln (w_{1} T_{1, 0} + w_{2} T_{2, 0}) - w_{1} ln T_{1, 0} - w_{2} ln T_{2, 0} \geq 0,

(7)

with equality only for equal initial temperatures. In this proof all temperatures inside the logarithms mean their numerical values in one common absolute unit. This is Jensen's inequality applied to the same weighted average that conservation of energy selected. The direction of heat transfer and the stability of its endpoint also follow from entropy maximization under thermal contact.

Heat bath relaxation with finite capacity ( $τ = C_{1} C_{2} / [G (C_{1} + C_{2})]$ )

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 are

\dot{Q} = G (T_{1} - T_{2}), C_{1} \dot{T}_{1} = - G (T_{1} - T_{2}), C_{2} \dot{T}_{2} = G (T_{1} - T_{2}) .

The 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 gives

\frac{d}{d t} (T_{1} - T_{2}) = - G (\frac{1}{C _{1}} + \frac{1}{C _{2}}) (T_{1} - T_{2}), τ = \frac{C _{1} C _{2}}{G ( C _{1} + C _{2} )} .

The temperature difference has exponential decay with time constant

τ

. For the finite heat bath example and the stipulated

G = 2 W K^{- 1} = 2 J s^{- 1} K^{- 1}

τ = 45 s

, and solving the ordinary differential equations gives

T_{1} (t) = 296 K + 54 K e^{- t / (45 s)}, T_{2} (t) = 296 K - 6 K e^{- t / (45 s)},

Q (t) = 5400 J (1 - e^{- t / (45 s)}) .

At one time constant,

T_{1} ≃ 315.865 K

T_{2} ≃ 293.793 K

and

Q ≃ 3413.451 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.

Figure 1.
Finite bath and fixed bath Newton's law of cooling histories
. Original analytical model predictions with $C_{1} = 100 J K^{- 1}$ , $C_{2} = 900 J K^{- 1}$ and $G = 2 W K^{- 1}$ ; time is in seconds. Solid lines show the two finite bodies approaching $296 K$ with time constant $45 s$ . Dashed lines show body 1 cooling toward a bath held at $290 K$ , with time constant $50 s$ . The two panels use different vertical scales so that the finite bath's $6 K$ warming remains visible. **Details**.

For an ideal heat bath with the same

G

, set

T_{2} = T_{b} = 290 K

rather than evolving its temperature. Newton's law of cooling then yields

T_{1} (t) = 290 K + 60 K e^{- t / (50 s)}, τ_{\infty} = C_{1} / G = 50 s .

Changing

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:

\dot{S}_{tot} = \frac{- Q ˙}{T _{1}} + \frac{Q ˙}{T _{2}} = \frac{G ( T _{1} - T _{2} ) ^{2}}{T _{1} T _{2}} \geq 0.

(6)

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.

Regulated calibration bath

A regulated bath holds its temperature near a chosen setting through feedback control, while stirring reduces internal temperature gradients. Its finite heat capacity buffers disturbances; external heating and cooling replace the ideal heat bath's unlimited capacity. This is useful for thermometer calibration and for maintaining a blackbody calibration source at a known temperature.

Joel B. Fowler's A Third Generation Water Bath Based Blackbody Source reports a NIST test of a stirred water bath containing a copper blackbody cavity. A bottom plate supplied heating and cooling; a double stirrer circulated water. Electronic feedback control regulated heating. An electronic thermometer with a thermistor probe provided the readout. The preparation was to approach a selected setting, wait for stabilization, adjust against the measured temperature and then log continuously.^[ref]

To test spatial uniformity, one thermistor stayed near the cavity tip while another visited ten locations at three heights. The simultaneous reference distinguished spatial differences from temporal drift. An insulated dome reduced thermal conduction along probe casings. Table 1 reports nonzero departures at several settings: circulation did not make every location identical. The measurement uncertainty analysis separately included uniformity, stability, setting, immersion and readout contributions. These tests supported a bounded temperature assignment around the cavity, rather than infinite heat capacity or perfect thermal equilibrium throughout the apparatus.^[ref]

Entropy maximization under thermal contact ( $\partial S_{tot} / \partial E_{1} = 1/ T_{1} - 1/ T_{2}$ )

At fixed volumes and material inventories, two internally equilibrated bodies that exchange only heat within an isolated system have fixed total internal energy

E = E_{1} + E_{2}

. Their total entropy is

S_{tot} (E_{1}) = S_{1} (E_{1}) + S_{2} (E - E_{1}) .

Using

1/ T_{i} = (\partial S_{i} / \partial E_{i})_{V_{i}, N_{i}}

, an interior stationary point obeys

\frac{d S _{tot}}{d E _{1}} = \frac{1}{T _{1}} - \frac{1}{T _{2}} = 0.

Thus entropy stationarity requires matching thermodynamic temperatures. Positive values of heat capacity at constant volume make this a strict local maximum because

\frac{d ^{2} S _{tot}}{d E _{1}^{2}} = - \frac{1}{C _{1} T _{1}^{2}} - \frac{1}{C _{2} T _{2}^{2}} < 0.

For bodies with positive heat capacities throughout the accessible range, this concavity gives at most one interior maximum. This is the stability check behind the equal-temperature condition; see entropy concavity and positive heat capacity. ^[ref]

For thermal equilibration of two finite bodies, entropy from heat capacity gives

Δ S_{tot} = C_{1} ln \frac{T _{f}}{T _{1, 0}} + C_{2} ln \frac{T _{f}}{T _{2, 0}} = (ln \frac{375}{300} + 3 ln \frac{375}{400}) kJ K^{- 1} ≃ 0.02952799 kJ K^{- 1} > 0.

Body 1 gains about

0.22314355 kJ K^{- 1}

of entropy, while body 2 loses about

0.19361556 kJ K^{- 1}

. Total energy is conserved even though total entropy increases. The entropy calculation uses reversible comparison paths between the endpoint states; actual exchange across a finite temperature difference is irreversible.

Figure 1.
At fixed total energy, matching temperatures maximize total entropy
. The curve is total entropy relative to the initial $300 K$ , $400 K$ state. Every point satisfies $C_{1} T_{1} + C_{2} T_{2} = 1500 kJ$ . Only $250 \leq T_{1} / K \leq 500$ is shown; the mathematical model's positive-temperature domain is $0 < T_{1} / K < 1500$ . **Details**.

To construct the plot, vary

T_{1}

while enforcing the energy constraint, rather than holding

T_{2}

fixed. In numerical units of kelvin,

T_{2} / K = (1500 - T_{1} / K) /3

, and the plotted function is

S (T_{1}) = C_{1} ln \frac{T _{1}}{300 K} + C_{2} ln \frac{T _{2} ( T _{1} )}{400 K}, \frac{d T _{2}}{d T _{1}} = - \frac{C _{1}}{C _{2}} .