According to Wikipedia, the enthalpy of formation of water is $-285.8~\mathrm{kJ/mol}$ while the enthalpy of formation of steam is $-241.818~\mathrm{kJ/mol}$, implying the following:
$\ce{H2O(l) -> H2O(g)}\qquad (\Delta H=44.0~\mathrm{kJ/mol)}$
Let us derive the enthalpy of the above reaction using the specific heat capacity of water and the specific heat of vaporization of water instead.
$1~\mathrm{mol}$ of water weighs $0.018~\mathrm{kg}$.
To raise $1~\mathrm{mol}$ of water from $25~\mathrm{^\circ C}$ to its boiling temperature requires $(0.018~\mathrm{kg}) \times (4200~\mathrm{J~kg^{-1}~K^{-1}}) \times (75~^\circ\mathrm{C}) = 5.67~\mathrm{kJ}$.
To turn that amount of water to steam requires $(0.018~\mathrm{kg}) \times (2258~\mathrm{kJ~kg^{-1}}) = 40.644~\mathrm{kJ}$.
Adding these two terms give $46.314~\mathrm{kJ}$, implying the following:
$\ce{H2O(l) -> H2O(g)}\qquad (\Delta H=46.314~\mathrm{kJ/mol)}$
Is the discrepancies between the two thermochemical equations simply due to measurement inaccuracies?
