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 Incoming links: Reversible heat engine

C_{1}, C_{2}

, a reversible heat engine taking two bodies to a common temperature has

T_{f} = T_{1}^{C_{1} / (C_{1} + C_{2})} T_{2}^{C_{2} / (C_{1} + C_{2})}

. The maximum work is

C_{1} T_{1} + C_{2} T_{2} - (C_{1} + C_{2}) T_{f}

. heat isolation does not prohibit transferring work to a store included in the total isolated system.

To obtain the maximum work from two finite heat reservoirs, operate a reversible heat engine between the bodies until their temperatures coincide. Include the work store in the isolated total system: no heat enters from outside, while the bodies can transfer energy to that store. With the stated constant heat capacities, reversibility requires

C_{1} ln \frac{T _{f}}{T _{1}} + C_{2} ln \frac{T _{f}}{T _{2}} = 0, T_{f} = T_{1}^{C_{1} / (C_{1} + C_{2})} T_{2}^{C_{2} / (C_{1} + C_{2})} .

Numerically,

T_{f} = 45 0^{5/8} 27 3^{3/8} = 373.094 K

. The decrease of the bodies' internal energy is

W_{m a x} = C_{1} T_{1} + C_{2} T_{2} - (C_{1} + C_{2}) T_{f} = 2.1061 \times 1 0^{4} J ≃ 21.1 kJ .

Direct thermal equilibration without an engine would instead reach the arithmetic heat-capacity-weighted temperature

383.625 K

and deliver no work. entropy production can only raise the final common temperature above the value for zero entropy production and reduce extractable work.