---
title: "Mixing 12 kg water and 8 kg ice: entropy change"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:water-ice-mixing-adiabatic-entropy"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/water-ice-mixing-adiabatic-entropy
site: BCT Engineering
---

# Mixing 12 kg water and 8 kg ice: entropy change

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

- Asked in: 2070 Magh, 8 marks
- Video: [AP Physics 2 Thermodynamics - Entropy Example (Mixing Water) (Again?)](https://www.youtube.com/watch?v=clO_HKj7Eb8) — Coach Botzer
- 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/water-ice-mixing-adiabatic-entropy
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

A control mass system consists of ice and water: 12 kg of water, at 37°C is mixed with 8 kg of ice at -27°C. Assuming the process of mixing is adiabatic, find the change of entropy. Latent heat of ice = 336 kJ/kg, Cp for water = 4.2 kJ/kg·K.

## Short answer

The total entropy change of the adiabatic mixing is 0.51 kJ/K. The final mixture is water and ice at 0°C, with 4.2 kg of ice melted.

## Answer

**Missing data.** The specific heat of ice is not given; the usual value cp,ice = 2.1 kJ/kg·K is assumed.

**Given:** water m_w = 12 kg at 37 °C (310.15 K), c_w = 4.2 kJ/kg·K; ice m_i = 8 kg at −27 °C (246.15 K); L = 336 kJ/kg. The mixing is adiabatic.

#### Step 1: Find the final state
- Heat released if all the water cools to 0 °C: 12 × 4.2 × 37 = 1864.8 kJ
- Heat needed to warm the ice to 0 °C: 8 × 2.1 × 27 = 453.6 kJ
- Heat needed to melt all the ice: 8 × 336 = 2688 kJ

1864.8 > 453.6 but 1864.8 < 453.6 + 2688 = 3141.6. So the ice reaches 0 °C but melts only partly. The final state is a water-ice mixture at **0 °C (273.15 K)**.

Mass of ice melted: m_melt = (1864.8 − 453.6)/336 = **4.2 kg**. The final mixture has 16.2 kg water and 3.8 kg ice.

#### Step 2: Entropy changes
- Water cooling 37 °C to 0 °C: ΔS_w = 12 × 4.2 × ln(273.15/310.15) = **-6.4026 kJ/K**
- Ice warming −27 °C to 0 °C: ΔS_i1 = 8 × 2.1 × ln(273.15/246.15) = **1.7485 kJ/K**
- Melting 4.2 kg of ice at 273.15 K: ΔS_i2 = 4.2 × 336/273.15 = **5.1664 kJ/K**

#### Step 3: Total
ΔS = -6.4026 + 1.7485 + 5.1664

**Answer: ΔS = 0.512 kJ/K ≈ 0.51 kJ/K** (positive, as the mixing is irreversible and adiabatic).

## Check yourself

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

1. What assumption is made for the missing data?
   A. Specific heat of ice = 2.1 kJ/kg·K
   B. Ice melts fully
   C. Latent heat = 0
   D. Final T = 37 °C
   Answer: A. Typical value cp,ice = 2.1 kJ/kg·K is assumed. (Key point: Missing data)

2. The final state is a water-ice mixture at 0 °C because:
   A. the water heat available (1864.8 kJ) exceeds that to warm the ice but is less than warm plus melt (3141.6 kJ)
   B. the ice is colder than water
   C. the water cools below 0 °C
   D. all ice melts
   Answer: A. Heat from water cannot melt all the ice, so some stays ice. (Key point: Final state)

3. Mass of ice melted is:
   A. 8 kg
   B. 5.55 kg
   C. 3.8 kg
   D. 4.2 kg
   Answer: D. (1864.8 − 453.6)/336 = 4.2 kg; 3.8 kg stays ice. (Key point: Mass melted)

4. Entropy change total (−6.4026 + 1.7485 + 5.1664) is:
   A. +0.51 kJ/K, positive
   B. −0.51 kJ/K
   C. 0
   D. −13.3 kJ/K
   Answer: A. The sum is positive, consistent with an irreversible adiabatic mixing. (Key point: Total entropy)

## Related in Fundamental of Thermodynamics & Heat Transfer

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