Enthalpy of Water and Energy Required to obtain a Quality X in a water-vapor mixture
Between 20 degrees Celsius and 100 degrees Celsius the average values of density (rho) and specific
heat (Cp) of liquid water are
rho = 981.2 kg/m^3
Cp = 4191.5 J/kg C
So that the Enthalpy of liquid water as a function of temperature (using T=0 C as reference
temperature) is
H_L = rho*Cp*T = 4,112,699 * T J/m^3
At the boiling point (Tb = 100 C) , the value of the enthalpy is thus
H_L(@Tb) = 411,269,900 J/m^3
The latent heat of vaporization of water is
H_vap = 2,227,324,000 J/m^3
Consider a container containing 1 m^3 of water at T = 0 C. The amount of energy required to heat the
water just to the boiling point (T = 100 C; quality X=0) is then 411,269,900 J.
Now consider the same 1 m^3 of water that has been heated just to the boiling point (T = 100 C; quality
X=0). The amount of energy required to completely vaporize the water (T=100 C, quality X=1) is then
2,227,324,000 J.
The following table gives the amounts of energy required to produce water-vapor mixtures of various
qualities starting from 1 m^3 of liquid water at the boiling temperature and with quality of zero
(assuming that quantity of vapor formed is directly proportional to the energy input).
Energy Input (J)
0
Quality (-)
0
222,732,400
0.1
445,464,800
0.2
668,197,200
0.3
890,929,600
0.4
1,113,662,000
0.5
Therefore, a total of 1,113,662,000 J of energy are required to convert the liquid water at the boiling
point into a 50-50 mixture of liquid water and vapor at the boiling point.
Note that if one were to start with 1 m^3 of water at T=0, and additional 411,269,900 J of energy would
be required to be added to reach the same quality levels in the table above.