Question

A 5-kg steel container is cured at 500°C. An amount of liquid water at 15°C, 100 kPa is added to the container so a final uniform temperature of the steel and the water becomes 75°C. Neglect any water that might evaporate during the process and any air in the container. How much water should be added and how much entropy was generated?

Answer 1

Heat steel 

 m(u₂ − u₁) = ₁Q₂ = mC (T₂ + T₄ )

₁Q₂ = 5(0.46)(500-20) = 1104 kJ 

 m_H2O( u₃-u₂)H₂O + m_st( u₃-u₂) = 0

m_H2O( 313.87 – 62.98) + m_stC ( T_3-T₂) = 0

m_H2O 250.89 + 5 × 0.46 × (75 - 500) = 0

m_H2O = 977.5/250.89 = 3.896 kg

m_H2O ( s₃-s₂) + m_st( s₃ - s₂) = ∅ + ₂S₃ gen

3.986 (1.0154 – 0.2245) + 5 × 0.46 ln 75+273/773 = ₂S₃ gen

₂S₃ gen = 3.0813 – 1.8356 = 1.246 kJ/K