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?
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.
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. :-)
Your calculations for the heat absorbed during each phase change are correct. However, the sign conventions for the heat calculations need to be adjusted.
Let's go through the calculations step by step:
1. From 117°C to 100°C:
Here, the steam is losing heat, so the heat absorbed should be negative:
q = (58 g)(2.03 J/g°C)(-17°C) = -1983.74 J
2. Heat of Condensation:
Here, the steam is condensing into liquid water, so the heat released should also be negative:
q = (58 g)(-2260 J/g) = -130780 J
3. From 100°C to 0°C:
Here, the liquid water is losing heat, so the heat absorbed should again be negative:
q = (58 g)(4.18 J/g°C)(-100°C) = -24244 J
4. Heat of Solidification:
Here, the liquid water is freezing into ice, so the heat released should be negative:
q = (58 g)(-334 J/g) = -19372 J
5. From 0°C to -32.5°C:
Here, the ice is losing heat, so the heat absorbed should be negative:
q = (58 g)(2.09 J/g°C)(-32.5°C) = -3906.55 J
Now, let's add up all the heats:
q_total = -1983.74 J + (-130780 J) + (-24244 J) + (-19372 J) + (-3906.55 J)
= -176286.29 J
To convert the answer to kilojoules, divide by 1000:
q_total = -176286.29 J / 1000 = -176.28629 kJ
Therefore, the correct amount of heat absorbed when 58 g of steam at 117°C are completely converted to ice at -32.5°C is -176.28629 kJ.
It appears that your answer of -180.63723 kJ was very close but slightly off. Make sure you correctly account for the signs of heat absorbed and released during the calculations.