Discrepancies between specific heat of vaporization of water and enthalpy of formation of water and steam
According to “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL, the standard molar enthalpy of formation at a temperature of $T=298.15\ \mathrm K$ is

$\Delta_\mathrm fH_\mathrm m^\circ=-285.8\ \mathrm{kJ\ mol^{-1}}$ for liquid water,

and

$\Delta_\mathrm fH_\mathrm m^\circ=-241.8\ \mathrm{kJ\ mol^{-1}}$ for gaseous water.

(These values are in agreement with the values that you have found on Wikipedia.)

Both values refer to a temperature of $T=298.15\ \mathrm K$; therefore, you do not have to raise the temperature of the liquid water to its boiling point. Nevertheless, you may do this; however, then you would have to cool down the steam back to $T=298.15\ \mathrm K$ later.

Instead, you can directly use the enthalpy of vaporization at the given temperature. Note that the enthalpy of vaporization depends on temperature, and that your value for the specifc enthalpy of vaporization of $\Delta_\text{vap}h=2258\ \mathrm{kJ\ kg^{-1}}$ refers to the boiling point at normal pressure, i.e. to a temperature of approximately $T=100\ \mathrm{^\circ C}$.

According to REFPROP – NIST Standard Reference Database 23, Version 9.0, the molar enthalpy of vaporization at a temperature of $T=298.15\ \mathrm K$ is

$\Delta_\text{vap}H_\mathrm m=43.987\ \mathrm{kJ\ mol^{-1}}$,

which is in good agreement with the difference of the values for the molar enthalpy of formation:

$-241.8\ \mathrm{kJ\ mol^{-1}}-\left(-285.8\ \mathrm{kJ\ mol^{-1}}\right)=44.0\ \mathrm{kJ\ mol^{-1}}$