Question

Calculate the amount of heat absorbed when 58 g of steam at 117◦C are completely converted to ice at −32.5◦C.

Specific Heat:
H2O(s) = 2.09 J/g◦C
H2O(l) = 4.18 J/g◦C
H2O(g) = 2.03 J/g◦C

Heat of Fusion = 334 J/g
Heat of Vaporization = 2260 J/g
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I got -180.63723 kJ, but it said I got the answer wrong.
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Work:
From 117◦C to 100◦C: q=(58 g)(2.03 J/g◦C)(-17◦C)=-2001.58 J
Heat of Condensation: q=(58 g)(-2260 J/g)=-131080 J
From 100◦C to 0◦C: q=(58 g)(4.18 J/g◦C)(-100◦C)=-24244 J
Heat of Solidification: q=(58 g)(-334 J/g)=-19372 J
From 0◦C to -32.5◦C: q=(58 g)(2.09 J/g◦C)(-32.5◦C)=-3939.65
I then added all of the heats and divided by 1000 to get the answer in units of kJ.

Did I do anything wrong?

Answer 1

You're method looks ok to me. I added the individual amounts and that looks ok to me. I did not check each individual calculation. I would look at two things. First, the heat is not absorbed; it is released when going from high T to lower T. And each step is a release and not an absorber. Second, it appears you have 3 significant figures and your answer contain many more than that. I don't know what to tell you about the sign but if I were making up the problem I would have asked how much heat was released then I would not have expected the - sign to be included. Second, you might try rounding that answer to 3 s.f. So I would remove the - sign and adjust the s.f.

Answer 2

Of course this could be a trick question. Since this is an exothermic transition the technical answer to the question is zero heat is absorbed. :-)