How much heat is required to vaporize 50.0 g of water if the initial temperature of the water is 25.0◦c and the water is heated to its boiling point where it is converted to steam? the specific heat capacity of water is 4.18 j · ( ◦c)−1 · g −1 and the standard enthalpy of vaporization of water at its boiling point is 40.7 kj · mol−1?

Q = mCΔT
Q is heat, m is mass, C is specific heat, and ΔT is change in temp

First, find how much energy you need to raise the heat to 100C, or the boiling point
Q = (50g)(4.18)(75 degrees) = 15675 J or 15.675 kJ

Then you need 40.7 kJ extra for each mole of water to evaporate it
If you have 50 grams of water and there are 18 grams per mol
50g / 18g = 2.8 moles x 40.7 = 113 kJ extra

15.675 + 113 kJ = 128.7 kJ required

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