[ILoveISO]

How much heat is required to heat 100g water from -20 to 120 degrees C? Given the specific heat of liquid water = 4.184 J/gdegreeC
Specific heat of solid water = 2.06 J/g degree C
Specific heat of water vapor = 1.84 J/g degree C
Change in Hfus = 6.020 Kj/mol
Change in Hvap = 40.7 Kj/mol


2.06 x 100g x 20 degrees = 4120 J

4.184 x 100g x 100degrees = 41,840 J

1.84 x 100g x 20degrees = 3680 J

6.020 x 1000J x 100g divide by 18.016 = 33,414J

40.7 x 1000J x 100g divide by 18.016 = 225,910 J

Add all up I get 309Kj of energy, anything wrong?