How much heat must be added to 20 grams of water at 100 degrees Celsius to convert it to steam?

Question

My work
Q = (20)(4.18 x 10^3)(100)
= 8360000 J
Is that the correct answer?

Answer 1

No, it is already at 100, the boiling point
You want

.020 kg * Heat of Vaporization of water in Joules/ kg
so

.020 kg * 2.26*10^6 J/kg