18.0 g of liquid water is initially at 0 C. How much heat, in kJ, must be added to transform this into 18.0 g of steam at 200 C? The heat capacity of liquid water is 75.3 J/mol C. The heat capacity of steam is 37.4 J/mol C. The heat of vaporization is 40.65 kJ/mol.

Answer 1

q1 = heat to raise T form zero C to 100 C.
q1 = mass H2O x specific heat water x (Tfinal-Tintial)

q2 = heat to vaporize liquid water at 100 C to steam at 100 C.
q2 = mass H2O x heat vaporization

q3 = heat to raise T of steam from 100 C. to 200 C.
q3 = mass steam x specific heat steam x (Tfinal-Tinitial).

Answer 2

Do I add all of the q's to get the final answer?
Also, the problem does not give the "specific heat" or "mass" of steam. It only gives heat capacity of steam and water.

Answer 3

Yes, total Q = sum of individual q1, q2 , q3.

You have given the specific heats called heat capacity. The only thing you need to be careful about is that the mass of water etc is given as 18 grams, the unit for heat capacity is in kJ/mol so you must convert mass H2O to mols (moles = grams/molar mass and this probably is the easy route to go) or convert heat capacity to J/gram. I would convert 18 g to moles by 18/molar mass H2O AND I would convert heat capacity to J/mol which means multiplying by 1000 to convert kJ/mol to J/mol.
Yes, you have the mass for steam. Don't you expect that 18.0 g H2O would give you 18.0 g ice when it freezes and 18.0 grams steam when it vaporizes?