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Entropy change on mixing ice and water

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.

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]

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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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