= Finite heat bath
{title2=$\Delta T_b=Q_b/C_b$}
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
$$
Q_b=\Delta U_b=\int_{T_{b,0}}^{T_{b,f}}C_b(T)\,dT.
$$
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; $C_bT_b$ 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> $C_1=100\,\mathrm{J\,K^{-1}}$ and $C_2=900\,\mathrm{J\,K^{-1}}$, 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> give
$$
C_1(T_f-T_{1,0})+C_2(T_f-T_{2,0})=0,\qquad
T_f=\frac{C_1T_{1,0}+C_2T_{2,0}}{C_1+C_2}=296\,\mathrm K.
$$
Thus the <heat> $Q$ transferred from body 1 to bath 2 is
$$
Q=C_1(350-296)\,\mathrm K=C_2(296-290)\,\mathrm K=5400\,\mathrm J.
$$
Body 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.
\Image[/finite-heat-bath-temperature-and-energy.png]
{title=Finite bath <temperature> changes and the integrated <heat> ledger}
{description=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 $Q=5400\,\mathrm J$ and $T_f=296\,\mathrm K$. Right: the shaded <integrals> $\int C_i\,dT$ have equal area $5400\,\mathrm J$, although their widths are $54\,\mathrm K$ and $6\,\mathrm K$. The outgoing <heat> magnitude is shown positive; body 1's signed <internal energy> change is negative.}
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> are
$$
T_f-T_{2,0}=\frac{C_1}{C_1+C_2}(T_{1,0}-T_{2,0}),\qquad
\lim_{C_2\to\infty}T_f=T_{2,0}=290\,\mathrm K.
$$
Hold $C_1$ and both initial <temperatures> fixed in this <limit>. The transferred <heat> then approaches $C_1(T_{1,0}-T_{2,0})=6000\,\mathrm J$. 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:
$$
\Delta S_1=C_1\ln\frac{296}{350}\simeq-16.75737\,\mathrm{J\,K^{-1}},\qquad
\Delta S_2=C_2\ln\frac{296}{290}\simeq18.43068\,\mathrm{J\,K^{-1}},
$$
$$
\Delta S_{\rm tot}=C_1\ln\frac{T_f}{T_{1,0}}+C_2\ln\frac{T_f}{T_{2,0}}
\simeq1.673308\,\mathrm{J\,K^{-1}}>0.
$$
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
$$
\frac{\Delta S_{\rm tot}}{C_1+C_2}
=\ln(w_1T_{1,0}+w_2T_{2,0})-w_1\ln T_{1,0}-w_2\ln T_{2,0}\ge0,
$$
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>.
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