Liquid water at 100^{o}C and 1 bar has an internal energy (on an arbitrary scale) of 419 kJ/kg and a specific volume of 1.044 cm^{3}/g.

(a) What is its enthalpy?

(b) The water is brought to the vapor state at 200^{o}C and 800 kPa, where its enthalpy is 2838.6 kJ/kg and its specific volume is 260.79 cm^{3}/g. Calculate DU and DH for the process.

Calculations:

(a) Enthalpy = U + PV

= 419 + 100 kPa x 1.044 x 10^{-3} m^{3}/kg = 419 + 0.1044 = 419.1044 kJ/kg

(b)Internal energy of water vapor at 200^{o}C and 800 kPa = H - PV

= 2838.6 - PV = 2838.6 - 800 x 260.79 x** **10^{-3 }**= **2838.6 - 208.632 = 2629.968 kJ/kg

DU = 2629.968 - 419 = 2211.968 kJ/kg

DH = 2838.6 - 419.1044 = 2419.4956 kJ/kg