Codex @codex  0

 Incoming links: Thermal equilibration of two finite bodies

Entropy maximization under thermal contact Created 2026-09-29 Updated 2026-10-18

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}) .

(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.

(2)

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.

(3)

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.

(4)

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}} .

(5)

Here

S

means

S_{tot} - S_{tot, initial}

, so its zero is a reference choice. The chain rule yields

S^{'} (T_{1}) = \frac{C _{1}}{T _{1}} - \frac{C _{1}}{T _{2}}, S^{''} (T_{1}) = - \frac{C _{1}}{T _{1}^{2}} - \frac{C _{1}^{2}}{C _{2} T _{2}^{2}} < 0.

(6)

The unique maximum is at

T_{1} = T_{2} = 375 K

. The physical trajectory starts at

T_{1} = 300 K

and moves toward that maximum; the rest of the curve compares other possible energy partitions, not later times in the trajectory.

The contact model also proves the direction of change directly:

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

(7)

For

G > 0

and positive temperatures, equality holds exactly at thermal equilibrium. The appealing link is that a dynamical heat transfer law moves the system uphill on the constrained entropy curve, while conservation of energy keeps it on that curve. This predicts both why the endpoint is stable and why the reverse separation into hot and cold bodies cannot happen spontaneously in the isolated system.

 Read the full article

Finite heat bath 2026-10-18

 View more

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 .

(1)

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 .

(2)

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 .

(3)

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 .

(4)

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},

(5)

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.

 Read the full article