Solve a finite-body balance with unequal capacities

Declare the distribution or system

Two finite bodies exchange heat inside an isolated fixed-volume boundary. Body A has total heat capacity CA(T)=2+0.01T J/K; body B has CB=4 J/K. Their initial temperatures TA and TB differ. There is no external heat, work or phase transition, so the sum of their internal-energy changes must vanish.

Derive the integral or response

Integrate each capacity from its own starting temperature to the common candidate Tf. The residual is 0.005Tf²+6Tf−E, where E=2TA+0.005TA²+4TB. An arithmetic average of TA and TB ignores both unequal capacities and their temperature dependence.

Audit limits and evidence

The residual derivative is 6+0.01Tf, positive over the supported interval. At the colder initial temperature the total change is negative; at the hotter initial temperature it is positive. Thus there is exactly one root between them. The positive quadratic root can be evaluated as 2E/[6+sqrt(36+0.02E)], avoiding subtraction of nearly equal positive numbers.

Predict a changed premise

Check each body energy separately and then their sum. The warmer body loses energy and the cooler gains it under this model. The final-state balance says nothing about how long equilibration takes, whether a real enclosure leaks heat or whether a material changes phase; those would need additional response and boundary evidence.

Worked original case

At TA=500 K,TB=250 K, E=3250 J in the chosen energy reference. The root is about 404.987562 K. Body A loses about 619.950248 J and body B gains the same amount. The arithmetic midpoint 375 K fails the energy balance.

Investigation

Bracket the root with the two starting temperatures, explain uniqueness from positive total capacity and compare a candidate midpoint residual with the solved residual. Change one initial temperature without changing the capacity law.

Changed-case practice

Try the changed case before feedback

Take TA=400 K and TB=200 K. Find equilibrium Tf in kelvin.

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 graphics are reproduced. These bounded CPU calculations do not claim actual gas handling, pressure-vessel work, thermal exposure or practical measurement. Public feedback is practice, not independently verified reasoning; retain original work, critique, revision and a newly assigned case in the course evidence workspace.