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Mixing 12 kg water and 8 kg ice: entropy change

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.

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.

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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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