40.0 grams of liquid at its boiling point is changed to vapor by the addition of 12800
joules of heat. What is the heat of
vaporization of the substance?

320J/g

Explanation:

Given parameters:

Mass of the liquid  = 40g

Amount of heat added  = 12800J

Unknown:

The heat of vaporization = ?

Solution:

The heat of vaporization can be related to heat added by the expression below;

           H  = mL

H is the heat added

m is the mass

L is the heat of vaporization

 Insert the parameters and solve;

          12800 = 40 x L

            L =   = 320J/g

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