General Chem Question anyone?

[sddat]

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I know this is the wrong place to put this question..but I'm pretty desperate and I've tried General Chem forums.

QUESTION:
How many grams of ice at -10.0 C can be melted by the condensation of 800. grams of steam initially at 110.0 C?
*Heat of Fusion: 6.01kj/mol
*Heat of Vapoization at 100C is 40.7 kj/mol
*Specific heat of steam is 1.84 J/g C
*Specific heat of ice is 2.09 J/g C

My problem is, I don't even know where to start without having the grams of ice? I'm sure it's not too difficult of an equation. It's just pretty confusing because this is the first time I have not had the grams.


Thanks in advance for any help.

 

[harjay]

start with the Specific Heat of Steam, you need to know how many Joules you will get from the cooling of the steam
1.84 J/gC * 800g * 10C = 14720 Joules

Next calculate how many Joules you'll get from condensation
40.7KJ/mol * 1mol/18.02g * 800g = 1806.9 KJ (not Joules)

The Total Number of Joules available: 1821.6KJ

Next calculate how many grams of ice that can be heated AND melted, using X grams. (note the difference between J and KJ)
(Xg * 2.09J/gC * 10C) + (Xg * 6.01KJ/mol * 1mol/18.02g) = 20.9XJ + 0.334XKJ = The total Number of Joules provided from above = 1821.6J

I converted here to grams for easily calculations \/
0.0209XkJ + 0.334XkJ = 1821.6KJ
Solve for X
=5140 Grams
Does it make logical sense? Sure, you get 40.7 kj/mol for condensing the steam, and only 6kj/mol for melting ice. Therefore you will have much more ice than steam.

 

[sddat]

Thank you sooo much. I didn't expect someone to give me that much information on the problem. I'm going to study that problem over and over before the final. Again, thank you!!