---
title: "Reversible reservoirs; entropy change expressions"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:reversible-reservoirs-entropy-change"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/reversible-reservoirs-entropy-change
site: BCT Engineering
---

# Reversible reservoirs; entropy change expressions

Fundamental of Thermodynamics & Heat Transfer · Chapter 5: Second Law of Thermodynamics

- Asked in: 2073 Magh, 6 marks; 2070 Bhadra, 6 marks; 2067 Mangsir, 8 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/reversible-reservoirs-entropy-change
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

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.

## 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.

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A reversible heat transfer reservoir is best described as:
   A. a system exchanging only work
   B. a small system whose temperature changes with heat
   C. a large system exchanging only heat at constant temperature without internal irreversibility
   D. an insulated rigid tank
   Answer: C. Examples are the atmosphere, a lake or a furnace; its temperature stays constant. (Key point: Heat reservoir)

2. A reversible work transfer reservoir exchanges energy:
   A. only as work, with no heat and no dissipation
   B. as heat at constant T
   C. as both heat and mass
   D. only as heat
   Answer: A. Examples: frictionless pulley-weight, spring, flywheel; its entropy stays constant. (Key point: Work reservoir)

3. A reservoir at T_H loses heat Q reversibly. Its entropy change is:
   A. +Q/T_H
   B. 0
   C. −Q/T_H
   D. Q T_H
   Answer: C. ΔS = Q/T with Q negative when heat leaves, so ΔS = −Q/T_H. (Key point: Reservoir entropy change)

4. Reversible work transfer changes the entropy by:
   A. δW/T
   B. zero, so it is isentropic
   C. a positive amount
   D. a negative amount
   Answer: B. δQ = 0 and no dissipation give dS = 0. (Key point: Work carries no entropy)

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