How many kJ are required to heat 45.0 g of H2O at 25.0°C and then boil it all away?

The answer is #"116 kJ"#.

You need to go through two stages in order to find the total heat required to completely boil #"45.0 g"# of water. First, you must provide enough heat to get the water to #"100"^@"C"#. This is calculated by using

#q_1 = m * c * DeltaT#, where

#c# - water's specific heat - #"4.184 J/g" * ^@"C"#.

You get

#q_1 = "45.0 g" * 4.184 J/(g*^@C) * (100.0 -25.0)^@C = 14121# #"J"#

Then, you must provide enough heat to get all the water from liquid to steam. This is calculated using water's enthalpy of vaporization, #DeltaH_("vap")# - #"40.68 kJ/mol"#.

We can go to moles of water using its molar mass, #"18.0 g/mol"#

#n_("water") = m/("molar mass") = ("45.0 g")/("18.0" ("g")/("mol")) = 2.50# #"moles"#

#q_2 = n_("water") * DeltaH_("vap") = "2.50 moles" * "40.68" ("kJ")/("mol") = 101.7# #"kJ"#

Therefore, the total energy in the form of heat that is needed is

#q_("TOTAL") = q_1 + q_2 = 14121*10^(-3)# #"kJ" + 101.7# #"kJ"#

#q_("TOTAL") = 116# #"kJ"#