10 kg of water at 20°C is converted to ice at -10°C by being put in contact with a reservoir at -10°C. This process takes place at constant pressure. (a) Calculate the heat absorbed by the cold reservoir. (b) Calculate the change in entropy of the closed system “reservoir water/ice”.
Answer
Heat Absorbed and Entropy Change in Thermodynamic Process
🧾 Given Data
- Mass (m) = 10 kg
- Initial Temperature = 20°C = 293 K
- Final Temperature = –10°C = 263 K
- Latent Heat of Fusion (L) = 334 kJ/kg
- Specific Heat of Water (c₁) = 4.18 kJ/kg·K
- Specific Heat of Ice (c₂) = 2.1 kJ/kg·K
🔥 Step 1: Calculate Heat Absorbed by Cold Reservoir
We divide the process into 3 parts:
Q₁ = m × c₁ × (0 − 20) = 10 × 4.18 × (−20) = −836 kJ
Q₂ = −m × L = −10 × 334 = −3340 kJ
Q₃ = m × c₂ × (−10 − 0) = 10 × 2.1 × (−10) = −210 kJ
Total Heat Released: Q_total = −4386 kJ
Thus, Heat absorbed by the cold reservoir = 4386 kJ
♻️ Step 2: Entropy Change of the Closed System
Water from 20°C to 0°C:
ΔS₁ = 10 × 4.18 × ln(273/293) = −2.973 kJ/K
Freezing at 0°C:
ΔS₂ = −(10 × 334) / 273 = −12.24 kJ/K
Ice from 0°C to −10°C:
ΔS₃ = 10 × 2.1 × ln(263/273) = −0.777 kJ/K
Entropy gain by the reservoir:
ΔS_reservoir = 4386 / 263 = 16.68 kJ/K
Total Entropy Change:
ΔS_total = −2.973 − 12.24 − 0.777 + 16.68 = +0.69 kJ/K
✅ Final Answers
- Heat absorbed by cold reservoir: 4386 kJ
- Change in entropy of the closed system: +0.69 kJ/K
This positive change in entropy confirms consistency with the second law of thermodynamics.