Question

Determine the amount of heat (in kj) required to vaporize 1.55 kg of water at its boiling point. for water, δhvap

Answer 1

Heat required to vaporize 1 mol of water from water at 100C to steam at 100C = 40.7 kJ 
1 mol of water weighs =  18.015g
1.55 kg = 1550/18.015 mol = 86.03 mol 

Heat required to vaporize :
= 86.03 mol  x 40.7 kJ 

= 3501.421 kJ