Question

Two kg water at 200 kPa with a quality of 25% has its temperature raised 20°C in a constant pressure process. What is the change in enthalpy?

Answer 1

State 1 at 200 kPa

h = h_f + x h_fg = 504.68 + 0.25 × 2201.96 = 1055.2 kJ/kg

State 2 has same P at 200 kPa

T₂ = T_sat + 20 = 120.23 + 20 = 140.23°C

so state 2 is superheated vapor (x = undefined)