L = latent heat of vaporization for ethyl alcohol = 854 J/gram
m = mass of the ethyl alcohol to change from liquid to gas = 5.20 grams
Q = heat of vaporization required to change from liquid to gas = ?
heat of vaporization required to change from liquid to gas is given as
Q = m L
inserting the values
Q = (5.20 grams) (854 J/gram)
Q = (5.20 x 854) (J/gram) (gram)
Q = 4440.8 Joules
so heat of vaporization comes out to be 4440.8 Joules.