How Much Steam Is Needed to Heat Water and Melt Ice in a Calorimetry Experiment?

[Aeonic333]

My solution (179 g) to the problem below is slightly less than what the book says it should be (190 g)...

A vessel whose walls are thermally insulated contains 2.40 kg of water and 0.450 kg of ice, all at a temperature of 0.0 degrees Celsius. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.0 degrees Celsius? Neglect the heat transferred to the container.

I set the problem up like this...

Qw + Qi + Hf + Hv = 0

Then I split Qw into two separate methods of heat transfer, since there is condensed steam changing from 100 degrees Celsius to 28.0 degrees Celsius, and liquid water change from 0.0 degrees Celsius to 28.0 degrees Celsius. This seems logical to me, but according to the book I am wrong. Any suggestions?

Tim

 

[arcnets]

Hi Aeonic333,
in my naive understanding, the ice must melt before the temperature can rise. Did you take this into account?

 

[Aeonic333]

That is true. The ice must melt and the steam must condense, that is what Hf (heat of fusion) and Hv (heat of vaporization) stand for. I did make one mistake in my equation though: the heat of fusion needs to have a negative value since it is decreasing the overall energy of the system. So the new equation should be:

Qwater + Qice + Hv - Hf = 0

There still must be something I am neglecting, because I have yet to come up with the supposedly correct answer of 190 g for the mass of steam. I will be working on this until I figure it out, so please... SOMEBODY step up to the plate!