---
title: "Entropy change on mixing ice and water"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:ice-water-mixing-entropy-change"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/ice-water-mixing-entropy-change
site: BCT Engineering
---

# Entropy change on mixing ice and water

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

- Asked in: 2073 Magh, 8 marks
- Video: [Entropy and Mixing Ice with Water:  Example 2](https://www.youtube.com/watch?v=UO_i9k0rmb0) — Viken Kiledjian
- 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/ice-water-mixing-entropy-change
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

4 kg of water at 25°C is mixed with 1 kg of ice at 0°C in an isolated system. Calculate the change in entropy due to mixing process. [Take latent heat of ice L = 336 kJ/kg and specific heat of water c = 4.18 kJ/kg.K]

## Short answer

All ice melts; energy balance gives final temperature 3.923°C. Entropy changes: water cooling ΔS₁ = −1.226 kJ/K, ice melting ΔS₂ = 1.230 kJ/K, melt heating ΔS₃ = 0.060 kJ/K. Total entropy change ΔS = 0.064 kJ/K.

## Answer

**Given:** m_w = 4 kg at 25 °C (298.15 K), m_i = 1 kg ice at 0 °C (273.15 K), L = 336 kJ/kg, c = 4.18 kJ/kg·K. The system is isolated, so the heat lost by the water equals the heat gained by the ice.

#### Step 1: Will all the ice melt?
Heat available from cooling the water to 0 °C: 4 × 4.18 × 25 = 418 kJ
Heat needed to melt all the ice: 1 × 336 = 336 kJ

418 > 336, so all the ice melts and the final temperature is above 0 °C.

#### Step 2: Final temperature
Heat lost by water = heat to melt ice + heat to warm the melted ice from 0 °C to T_f:

4 × 4.18 × (25 − T_f) = 336 + 1 × 4.18 × T_f

418 − 16.72 T_f = 336 + 4.18 T_f
82 = 20.9 T_f

T_f = 3.923 °C = 277.07 K

#### Step 3: Entropy changes
- Water cooling: ΔS₁ = m c ln(T_f/T₁) = 4 × 4.18 × ln(277.07/298.15) = **-1.2258 kJ/K**
- Melting of ice at 273.15 K: ΔS₂ = mL/T = 336/273.15 = **1.2301 kJ/K**
- Melted water heated 0 °C to T_f: ΔS₃ = 1 × 4.18 × ln(277.07/273.15) = **0.0596 kJ/K**

#### Step 4: Total
ΔS = ΔS₁ + ΔS₂ + ΔS₃ = -1.2258 + 1.2301 + 0.0596

**Answer: ΔS = 0.0639 kJ/K ≈ 0.064 kJ/K** (positive, so the mixing is irreversible, as expected).

## Check yourself

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

1. How do we decide whether all ice melts?
   A. Compare heat released by cooling the water to 0 °C (418 kJ) with heat to melt the ice (336 kJ)
   B. Compare masses of water and ice
   C. Assume final T is 12.5 °C
   D. Compare 25 °C with 273 K
   Answer: A. 418 > 336 kJ, so all ice melts and the final temperature is above 0 °C. (Key point: Check whether ice melts fully)

2. The final temperature satisfies:
   A. 4 × 4.18 × (25 − T) = 336
   B. 4 × 4.18 × (25 − T) = 1 × 4.18 × T
   C. 4 × 4.18 × T = 336 + 1 × 4.18 × (25 − T)
   D. 4 × 4.18 × (25 − T) = 336 + 1 × 4.18 × T
   Answer: D. Water heat loss = latent heat of melting + heating the melt water from 0 °C; T_f = 3.92 °C. (Key point: Final temperature)

3. The entropy change of melting the ice is:
   A. mL/T at 273.15 K = 1.230 kJ/K
   B. mL/T_f
   C. mL × T
   D. m c ln(T_f/273.15)
   Answer: A. Melting occurs at constant 273.15 K, so ΔS = 336/273.15. (Key point: Entropy of melting)

4. The total ΔS is +0.064 kJ/K. This indicates:
   A. a reversible process
   B. an impossible process
   C. an irreversible process, as expected for an isolated system
   D. an error in calculation
   Answer: C. Positive ΔS for an isolated system means entropy was generated by finite-ΔT mixing. (Key point: Sign of result)

## Related in Fundamental of Thermodynamics & Heat Transfer

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- [Hydrogen in rigid vessel: heat, entropy change](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Ahydrogen-rigid-vessel-entropy-produced) — Fundamental of Thermodynamics & Heat Transfer, chapter 5