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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 givesIf 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 $C_1=100\mathrm{J}{\mathrm{K}}^{-1}$ and $C_2=900\mathrm{J}{\mathrm{K}}^{-1}$, with initial temperatures $T_{1,0}=350\mathrm{K}$ and $T_{2,0}=290\mathrm{K}$. 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 giveThus the heat $Q$ transferred from body 1 to bath 2 isBody 1 loses $5400\mathrm{J}$ of internal energy and bath 2 gains exactly that amount. Body 1 cools by $54\mathrm{K}$, while bath 2 warms by $6\mathrm{K}$. Equal heat transfers produce unequal temperature changes because the heat capacities differ.

The heat bath approximation can now be assessed against a chosen allowed temperature change $\delta T>0$. Constant $C_b$ requires $|Q_b|/C_b\le \delta T$, rather than merely calling the bath large. For this example, an allowed change of $0.1\mathrm{K}$ would fail because the computed change is $6\mathrm{K}$. The exact bath shift and its ideal-reservoir limit areHold $C_1$ and both initial temperatures fixed in this limit. The transferred heat then approaches $C_1(T_{1,0}-T_{2,0})=6000\mathrm{J}$. 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: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 giveswith 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.

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 distributionHere $E_n$ 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]

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.