---
title: "Entropy generation for a control mass"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:entropy-generation-control-mass"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/entropy-generation-control-mass
site: BCT Engineering
---

# Entropy generation for a control mass

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

- Asked in: 2069 Bhadra, 6 marks
- Video: [Entropy Balance | Thermodynamics | (Solved Examples)](https://www.youtube.com/watch?v=SUoSfmRtf1c) — Question Solutions
- 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-generation-control-mass
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

Derive expression of entropy generation for a control mass.

## Short answer

For a control mass, entropy generation is S_gen = S2 – S1 – ∫δQ/Tb ≥ 0. It is zero for reversible processes and positive for irreversible ones; for an adiabatic system, S2 – S1 = S_gen.

## Answer

A **control mass** is a closed system (fixed mass, no mass crossing the boundary). Entropy generation S_gen measures the entropy created inside it by irreversibilities.

#### Derivation
Consider a closed system that goes from state 1 to state 2 by an **irreversible** process A, with heat exchanged at boundary temperature T_b. It returns from 2 to 1 by a **reversible** process B. Together they form a cycle that is irreversible overall.

Clausius inequality: ∮ δQ/T ≤ 0 (strictly < 0 for an irreversible cycle).

```text
   P
   |  1 ______ A (irreversible)
   |   \          \
   |    \__ B ____2
   |   (reversible)
   +----------------> V
```

∫₁² (δQ/T_b)_A + ∫₂¹ (δQ/T)_B < 0

For the reversible path B, ∫₂¹ (δQ/T)_B = S₁ − S₂ (definition of entropy). So

∫₁² δQ/T_b + S₁ − S₂ < 0

S₂ − S₁ > ∫₁² δQ/T_b

Convert the inequality to an equation by adding a non-negative term S_gen:

**S₂ − S₁ = ∫₁² δQ/T_b + S_gen, with S_gen ≥ 0**

or in rate form

dS/dt = Σ Q̇_k/T_k + Ṡ_gen

#### Meaning of the terms
- S₂ − S₁ is the entropy change of the system.
- ∫ δQ/T_b is the entropy transferred by heat (positive, negative or zero).
- S_gen = S₂ − S₁ − ∫ δQ/T_b is the entropy generated:
  - S_gen = 0 for a reversible process
  - S_gen > 0 for an irreversible process
  - S_gen < 0 is impossible

For an adiabatic system (δQ = 0), S₂ − S₁ = S_gen ≥ 0. If the system rejects heat Q_out to surroundings at T₀, taking the boundary at T₀ gives S_gen = ΔS_system + Q_out/T₀, i.e. the sum of the entropy changes of the system and the surroundings.

## Check yourself

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

1. The derivation uses a cycle made of:
   A. two reversible paths
   B. an irreversible process 1-2 and a reversible return 2-1
   C. two isothermal processes
   D. an adiabatic process
   Answer: B. This cycle is irreversible overall, so Clausius gives ∮δQ/T < 0. (Key point: Derivation setup)

2. Which inequality results?
   A. S2 − S1 < ∫δQ/T_b
   B. S2 − S1 = ∫δQ/T_b
   C. S2 − S1 > ∫δQ/T_b
   D. S2 − S1 = 0
   Answer: C. ∫₁²δQ/T_b + S1 − S2 < 0. (Key point: Clausius inequality)

3. Adding S_gen gives:
   A. S2 − S1 = ∫δQ/T_b + S_gen, S_gen ≥ 0
   B. S2 − S1 = ∫δQ/T_b − S_gen
   C. S2 − S1 = S_gen − ∫δQ/T_b
   D. S2 − S1 = S_gen ≤ 0
   Answer: A. S_gen turns the inequality into an equation. (Key point: Final expression)

4. For an adiabatic closed system:
   A. ΔS = 0
   B. ΔS = S_gen ≥ 0
   C. ΔS < 0
   D. ΔS = ∫δQ/T
   Answer: B. δQ = 0, so the entropy change equals the entropy generated. (Key point: Adiabatic closed system)

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