The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 3.12 g of water boils at atmospheric pressure?

Answer 1

Assuming the 3.12 g of water is already at 100ºC, we can find the amount of heat needed to convert this to steam as follows:

1) convert 3.12 g H2O to moles:

3.12 g x 1 mol H2O / 18 g = 0.173 mols H2O

2) Find heat needed:

0.173 mols x 40.66 kJ/mol = 7.05 kJ of heat needed