Question

The molar heat of vaporization of water is 40.7kJ/mol. How much heat must be absorbed to convert 50.0 grams of liquid water at 100oC to steam at 100oC?
A) 1.46 X 10 kJ
B) 1.13 X 102kJ
C) 2.04 X 103kJ
D) 3.66 X 104kJ

Answer 1

During a phase change the temperature does not change since all of the heat is being absorbed in order to break the intermolecular forces.  Due to that, the formula will not need to have T in it and is actually q=nΔH(v).
n=the number of moles (in this case 2.778mol of water since you divide 50g by 18g/mol).
ΔH(v)=the molar heat of vaporization (in this case 40.7kJ/mol).
q=the heat that must be absorbed
q=2.778mol×40.7kJ/mol
q=113.1kJ
Therefore the water needs to absorb 1.13×10²kJ.

I hope this helps.  Let me know if anything is unclear.