>> Curious where you got the values for 44.01 kJ/mol <<
I think that should have been 44.1 kJ. I used standard heats of formation values of H2O (l) and H2O (g) from the appendix of one of my general chemistry textbooks. DHf° for H2O (l) = -285.85 kJ/mol and DHf° for H2O (g) = -241.8 kJ/mol. Therefore DH° for the reaction H2O(l) --> H2O (g) will be
DH° reaction = [(1 mol)(-241.8 kJ/mol)] - [(1 mol)(-285.85 kJ/mol) = 44.05 kJ
(This is just the "products minus reactants" method for calculating DH° of reactions using the standard enthalpies of formation of the substances.)
That is DH°vap at standard temperature, 25°C. But meanwhile I found some better values. According to one of my textbooks it is 44.94 kJ/mol at 0°C and 40.67 kJ/mol at 100°C, so it does vary with temperature somewhat. They are in the CRC Handbook also, under "Steam Tables" in my edition; at "T = 32°F, Evap = 1075.5 Btu/lbm" which converts to 45.07 kJ/mol). Here's a web site that has some of this information too: http://www.lsbu.ac.uk/water/data.html. It gives DHvap = 45.051 kJ/mol at 0°C. As you can see, the values vary a little from source to source. I'll just go with 45.05 kJ/mol.
>> If I used DH° sublimation that would just equal 44.01 kJ/mol for fusion + the 6.01 kJ/mol for vaporization? Hess' Law? <<
That's right, here's what you have:
H2O (s) --> H2O (l) DH = 6.01 kJ at 0°C
H2O (l) --> H2O (g) DH = 45.05 kJ at 0°C
H2O (s) --> H2O (g) DH = 51.06 kJ at 0°C
>> also where would I find the molecular mass of ice - would it be that much different than water? <<
Since ice is made of water, that's the molecular weight to use, 18.0153 g/mol!
>> So I need 7556.02 kJ of heat to melt the ball and then evaporate it. <<
That's right; it will be a little more if you use the higher DH of 51.06 kJ/mol.
Steve
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