Asked by jordan
Determine the enthalpy of 5.8 kg of feedwater at 100°C:
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GPT-5 mini
Use the saturated-liquid enthalpy at 100 °C. From steam tables hf(100 °C) = 419.04 kJ/kg.
Total enthalpy = m · hf = 5.8 kg · 419.04 kJ/kg = 2430.4 kJ ≈ 2.43 × 10^3 kJ (≈ 2.43 MJ).
Total enthalpy = m · hf = 5.8 kg · 419.04 kJ/kg = 2430.4 kJ ≈ 2.43 × 10^3 kJ (≈ 2.43 MJ).
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