Given 1 kg of water at 99.0°C and a very large block of ice at -1.0°C. A reversible heat engine absorbs heat from the water and expels heat to the ice until work can no longer be extracted from the system. At the completion of the process: (a) What is the temperature of the water? (b) How much heat has been absorbed by the block of ice in the process? (c) How much ice has been melted? (d) How much work has been done by the engine
Answer
Reversible Heat Engine: Water and Ice Thermodynamics Solution
🔷 Given Data:
- Mass of water, m₁ = 1 kg
- Initial temperature of water, T₁ = 99°C = 372 K
- Temperature of ice block, T₂ = –1°C = 272 K
- Specific heat of water, c = 4.2 kJ/kg·K
- Latent heat of fusion of ice, Lf = 334 kJ/kg
🔹 (a) Final Temperature of the Water
At the equilibrium point, the water and ice block reach the same temperature:
Final temperature = T₂ = 272 K or –1°C
🔹 (b) Heat Absorbed by the Ice
Heat released by water as it cools:
QH = m × c × (Tinitial − Tfinal)
QH = 1 × 4.2 × (372 − 272) = 420 kJ
Efficiency of the reversible engine:
η = 1 − (T₂ / T₁) = 1 − (272 / 372) ≈ 0.2688
Heat rejected to the cold reservoir (ice):
QC = QH × (1 − η) = 420 × 0.7312 ≈ 357.7 kJ
🔹 (c) Mass of Ice Melted
mice = QC / Lf = 357.7 / 334 ≈ 1.07 kg
🔹 (d) Work Done by the Engine
W = QH − QC = 420 − 357.7 = 62.3 kJ
✅ Final Answers Summary:
- (a) Final temperature of water = –1°C
- (b) Heat absorbed by ice = 357.7 kJ
- (c) Mass of ice melted = 1.07 kg
- (d) Work done by engine = 62.3 kJ