The Problem: 33.3 grams of ice at 0.00 °C has heat added to it until steam at 150.0 °C results. Calculate the total energy expended.
Solution
1) Melt
q = (33.3 g / 18.0 g mol¯1) (6.02 kJ / mol)
2) raise in temperature as a liquid
q = (33.3 g) (100.0 °C) (4.184 J/g °C)
3) Boil
q = (33.3 g / 18.0 g mol¯1) (40.7 kJ / mol)
4) raise in temperature as a gas
q = (33.3 g) (50.0 °C) (2.02 J/g °C)
The ChemTeam gets 103.7 kJ as the final answer.