Derive a work limit when both source temperatures change

Declare the system and evidence

Two finite bodies with constant total capacities Ch and Cc=4 J/K start at Th and 300 K. A cyclic reversible engine can operate between them in infinitesimal steps until their temperatures meet. Unlike ideal infinite reservoirs, these bodies cool and warm; their initial Carnot efficiency cannot be used unchanged for all later heat.

Derive the state or rate account

For the reversible combined process, entropy conservation requires Ch ln(Tf/Th)+Cc ln(Tf/300)=0. Thus Tf=exp[(Ch lnTh+Cc ln 300)/(Ch+Cc)], a capacity-weighted geometric mean. Work delivered equals initial minus final body energy: Wmax=ChTh+Cc 300−(Ch+Cc)Tf.

Audit the constraint and model

For a common final equilibrium with the same capacities and an irreversible cyclic engine, total entropy is nonnegative, so the final common temperature is at least the reversible value. Greater final body energy leaves less work. This is a constrained thermodynamic upper bound, not a claim that a real finite device reaches it or a claim about maximum power.

Predict a changed boundary

Direct isolated contact produces no useful work and ends at the weighted arithmetic mean. That hotter final state has positive entropy generation. For equal capacities 2 J/K and temperatures 800 K and 200 K, predict reversible Tf=400 K and Wmax=400 J; direct contact ends at 500 K. The distinction is not an energy-conservation failure: the engine exports energy as work.

Worked original case

For Ch=2 J/K, Th=600 K, Cc=4 J/K, Tc=300 K, reversible Tf=300×2^(1/3)≈377.976315 K. Maximum work is 2400−6Tf≈132.142110 J. Direct contact ends at 400 K and generates 2 ln(2/3)+4 ln(4/3)≈0.339798 J/K.

Investigation

Derive the weighted geometric and arithmetic endpoints, compare their energy and entropy accounts and justify why irreversible common endpoints cannot deliver more work under the same boundary.

Changed-case practice

Try the changed case before feedback

Two bodies each have C=2 J/K, initially 800 K and 200 K. Find reversible maximum work in joules.

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.