Separate entropy production from unequal body entropy changes

Declare the system and evidence

Two fixed-volume bodies form an isolated system. The initially hot body has total capacity CA(T)=2+0.01T J/K, while the cold body has CB=4 J/K. No external work, heat or phase transition is allowed. Energy, rather than an arithmetic temperature average, determines their common final equilibrium temperature.

Derive the state or rate account

Integrate capacities to obtain E=2Th+0.005Th²+4Tc. The final energy is 6Tf+0.005Tf². Its derivative 6+0.01Tf is positive, and the initial temperatures bracket the root. Use the stable expression Tf=2E/[6+sqrt(36+0.02E)] rather than subtracting nearly equal quadratic terms.

Audit the constraint and model

Entropy differences use separate reversible reference paths for each body: delta SA=2 ln(Tf/Th)+0.01(Tf−Th), and delta SB=4 ln(Tf/Tc). Their sum is positive for the supported unequal initial temperatures. This does not make the actual heat exchange reversible; the reference integrals evaluate endpoints while direct finite-temperature heat flow generates entropy.

Predict a changed boundary

The graph includes candidate common temperatures that do not conserve the original energy. They are not alternative accessible endpoints of this isolated system. Never choose the visually largest plotted entropy as the solution without enforcing the energy constraint. In a constrained equilibrium maximum, admissible variations preserve the system inventories, unlike this arbitrary common-temperature scan.

Worked original case

For Th=500 K, Tc=250 K, E=3250 J and Tf≈404.987562 K. The hot body loses about 1.371628 J/K, the cold gains about 1.929582 J/K and the isolated sum gains about 0.557954 J/K. These rounded entropy values should be checked from the displayed logarithms.

Investigation

Derive the unique energy root, integrate both capacity-over-temperature functions and retain their opposite signs. Reject a midpoint candidate using its energy residual; explain why arbitrary plotted candidates do not establish an entropy maximum.

Changed-case practice

Try the changed case before feedback

Two different constant-capacity bodies each have C=2 J/K and start at 400 K and 100 K. Direct isolated contact ends at 250 K. Find total delta S in J/K.

Numerical result in the requested unitsCompare this result

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original derivations and synthetic cases checked 31 October 2026 against the primary teaching reference. No source exercises or diagrams are reproduced. All thermal operations are mathematical proposals or supplied records, not physical heating, pressure-vessel or exposure instructions. Public feedback and retained prose do not certify independent reasoning or actual practical participation. Preserve first work, critique, revision and a new assigned response in the separate course workspace.