If + 201,140 Joules of energy was absorbed to vaporize water at 100° C, what was the mass (grams) of the water used?

Answer 1

q = m x ∆H_vap

q = heat = 201,140 J

m = mass = ?

∆H_vap = enthalpy of vaporization for water = 2260 J/g (you can look this up in a table)

Solving for m, we have...

m = 201,140 J / 2260 J/g

m = 89 g of water