Answer - Given, mass of water = 455 g , ti =
80.0oC, tf = 100.0oC
Specific heat of water = 4.184 J/goC , ∆Hvap = 40.7
kJ/mol
Moles of water = 455 g / 18.016 g.mol-1
                       Â
= 25.3 moles
We know the formula for the heat
q1 = m*C*∆t
   = 455 g * 4.184 J/goC *
(100.0-80.0)oC
   = 38074.4 J
Now heat from the 100.0 to 100.0oC
q2 = m*∆Hvap
     = 18518.5 J
So, total heat absorbed = q1 + q2
                                   Â
= 38074.4 + 18518.5
                                   Â
= 56593 J
                        Â
  = 56.59 kJ
