In order to convert from mass to moles, we need to divide its mass by the molecular weight, therefore the number of moles of \({{\rm{H}}_{\rm{2}}}{\rm{O}}\)is:
\({\rm{422\;g}}{{\rm{H}}_{\rm{2}}}{\rm{O \times }}\frac{{{\rm{1\;mol}}}}{{{\rm{18}}{\rm{.02\;gH}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{\rm{ = 23}}{\rm{.4\;mol}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)
Heat needed to bring this amount of water to the normal boiling point is:\(\begin{aligned}{c}{\rm{\Delta }}{{\rm{H}}_{\rm{1}}}{\rm{ = m}}{{\rm{C}}_{\rm{s}}}{\rm{\Delta T}}\\{\rm{ = (422\;g)}}\left( {{\rm{4}}{\rm{.184\;J/}}{{\rm{g}}^{\rm{^\circ }}}{\rm{C}}} \right){\rm{(100 - 23}}{\rm{.5)}}\\{\rm{ = 135,000Joule}}\end{aligned}\)
Heat needed to vaporize this amount of water:
\(\begin{aligned}{\rm{\Delta }}{{\rm{H}}_{\rm{2}}}{\rm{ = }}{{\rm{n}}_{\rm{\Delta }}}{{\rm{H}}_{{\rm{vap }}}}\\{\rm{ = (23}}{\rm{.4mol)(40,650\;J/mol)}}\\{\rm{ = 951,000Joule}}\end{aligned}\)
Heat needed to increase the temperature of the steam:
\(\begin{aligned}{\rm{\Delta }}{{\rm{H}}_{\rm{3}}}{\rm{ = m}}{{\rm{C}}_{\rm{s}}}{\rm{\Delta T}}\\{\rm{ = (422\;g)(2}}{\rm{.09\;J/g)(150 - 100)}}\\{\rm{ = 44}}{\rm{.100Joule}}\end{aligned}\)
Hence, Total heat required\({\rm{: - \Delta }}{{\rm{H}}_{\rm{1}}}{\rm{ + \Delta }}{{\rm{H}}_{\rm{2}}}{\rm{ + \Delta }}{{\rm{H}}_{\rm{3}}}{\rm{ = 1130\;kJ}}\)
