---
title: "Entropy change for reversible heat and work transfer"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:entropy-reversible-heat-work"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/entropy-reversible-heat-work
site: BCT Engineering
---

# Entropy change for reversible heat and work transfer

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

- Asked in: 2074 Bhadra, 6 marks; 2072 Magh, 6 marks; 2071 Bhadra, 6 marks (asked 3 times)
- Video: [Entropy Changes for Reversible Processes and The TdS equations](https://www.youtube.com/watch?v=j4gkeEUmvA4) — Christi Patton Luks
- 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/entropy-reversible-heat-work
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

Define entropy. Derive the expressions for change in entropy for reversible heat transfer and reversible work transfer process.

## Short answer

Entropy dS=(δQ/T)_rev, a property from Clausius inequality. Reversible heat transfer: dS=δQ_rev/T, ΔS=∫δQ_rev/T; at constant T ΔS=Q/T. Reversible work transfer: δQ=0 and no dissipation → dS=0, ΔS=0; work carries no entropy, process isentropic.

## Answer

**Entropy** is a property of a system that measures the molecular disorder (randomness) of the system. It is defined for a reversible process by dS = (δQ/T)_rev, and its unit is kJ/K. For any process, the entropy of an isolated system never decreases.

#### Entropy as a property (Clausius)
For any cycle, Clausius inequality gives ∮ δQ/T ≤ 0, with equality for a reversible cycle. Take two reversible paths A and B between states 1 and 2. The cycle 1-A-2-B-1 is reversible, so

∮ δQ/T = ∫₁² (δQ/T)_A − ∫₁² (δQ/T)_B = 0

Hence ∫(δQ/T)_rev is the same for every reversible path. It depends only on the end states, so it is a property, called entropy:

dS = (δQ/T)_rev, S₂ − S₁ = ∫₁² (δQ/T)_rev

#### Reversible heat transfer
A reversible heat transfer takes place across an infinitesimal temperature difference, at the temperature T of the reservoir or system. Then

dS = δQ_rev/T

- For heat added (δQ > 0), entropy increases.
- For heat removed (δQ < 0), entropy decreases.
- At constant temperature T: ΔS = Q/T.

Example: a reservoir at 500 K supplying 100 kJ reversibly: ΔS = −100/500 = −0.2 kJ/K.

#### Reversible work transfer
In a reversible work transfer there is no heat transfer (δQ = 0) and no friction or dissipation (as in a frictionless piston, spring or flywheel). Then

dS = δQ_rev/T = 0, so ΔS = 0

Work carries no entropy. A reversible work transfer process is isentropic (S = constant). If friction or any irreversibility is present, entropy is generated, so ΔS > 0.

**Summary:** dS = δQ_rev/T for heat transfer, and dS = 0 for reversible work transfer.

## Check yourself

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

1. Entropy for a reversible process is defined by:
   A. dS = δW_rev/T
   B. dS = (δQ/T)_rev
   C. dS = δQ·T
   D. dS = dU/P
   Answer: B. Entropy is the property whose change equals the reversible heat divided by the absolute temperature. (Key point: Definition of entropy)

2. Why is ∫(δQ/T)_rev a property?
   A. Because the temperature is constant
   B. Because it equals the work done
   C. Because heat is a point function
   D. Because it is the same for all reversible paths between two states, so it depends only on end states
   Answer: D. Clausius' ∮δQ/T = 0 for reversible cycles implies path independence. (Key point: Why entropy is a property)

3. For reversible heat transfer at constant T, ΔS is:
   A. Q/T, positive for heat added
   B. Q·T
   C. 0
   D. Q/T, always positive
   Answer: A. dS = δQ_rev/T; the sign follows the direction of heat flow. (Key point: Reversible heat transfer)

4. For a reversible work transfer process, the entropy change is:
   A. positive due to friction
   B. negative as work leaves
   C. W/T
   D. zero, as no heat flows and there is no dissipation
   Answer: D. δQ = 0 and no irreversibility, so dS = 0 (isentropic); work carries no entropy. (Key point: Reversible work transfer)

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