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Reversible reservoirs; entropy change expressions

Short answer

Reversible heat reservoir: large system at constant T exchanging only heat; ΔS=Q/T. Reversible work reservoir: exchanges only work, no heat, no dissipation, ΔS=0. Entropy property dS=(δQ/T)_rev yields for reversible heat transfer dS=δQ_rev/T, for reversible work transfer dS=0.

Define reversible heat transfer reservoir and reversible work transfer reservoir. Also derive the expressions for change in entropy due to reversible heat transfer and reversible work transfer processes.

Reversible heat transfer reservoir

A reversible heat transfer reservoir (thermal energy reservoir) is a very large system that exchanges energy only as heat, and whose temperature stays constant however much heat is added or removed. It is internally reversible: all the heat transfer takes place without friction or other irreversibility inside it. Examples: the atmosphere, a lake, a furnace.

For a reservoir at temperature T supplying or receiving heat Q, the entropy change is ΔS = Q/T (Q positive when heat enters it).

Reversible work transfer reservoir

A reversible work transfer reservoir (mechanical energy reservoir) is a system that exchanges energy only as work, with no heat transfer and no dissipation. Examples: a frictionless weight-and-pulley arrangement, a spring, a flywheel, a frictionless piston raising a weight. Its entropy stays constant.

Entropy as a property

By Clausius, ∮ (δQ/T)_rev = 0 for a reversible cycle. The integral ∫(δQ/T)_rev between two states is therefore path independent. This defines the property entropy:

dS = (δQ/T)_rev

Entropy change for reversible heat transfer

dS = δQ_rev/T, so S₂ − S₁ = ∫₁² δQ_rev/T.

For a reservoir at constant T: ΔS_reservoir = Q/T. When heat Q leaves a reservoir at T_H, ΔS = −Q/T_H. When it enters a reservoir at T_L, ΔS = +Q/T_L.

Entropy change for reversible work transfer

For reversible work, δQ = 0 and no dissipation occurs, so

dS = 0, ΔS = 0

The work reservoir undergoes an isentropic process. Work therefore transfers no entropy, while heat does.

Result: heat transfer changes entropy by δQ_rev/T; reversible work transfer changes it by zero.

✓ Answer checked by Claude Opus 5.5 (high effort)

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