Heat Energy Question

[caitie16]

If 0.5kg of steam at 100C is blown onto a block of ice at 0C, what will be the mass of the resulting water if all the ice does not melt?

Im not sure where to start with this. I tried:
ML + MC(delta)T = ML
but i keep getting the wrong answer. The answer is 4.5kg.
Could someone help me with this?

[caitie16]

I guess it kind of looks like i want someone to do my homework for me, but this isn't homework, its for my own understanding (which is somewhat lacking right now). My biggest problem is figuring out what the final temperature is.

I've tried this:

ML + MC(delta)T = ML + MC(delta)T

Heat energy of 0.5 kg of steam (water) = Heat energy of ice (water)

and I've tried making the final temp = 75C (because it was in previous similar question)
but obviously it doesn't work.

[mr_homm]

Hi catie16,

Since they say that not all the ice melts, you can picture it as a very large block with steam blowing down onto it from above, which will melt a hole in the top, forming a small lake of meltwater. Remember that when a material is at its melting or boiling point, its temperature cannot change, because any heat will go into changing its phase instead of its temperature. Therefore, the ice stays T=0C, and the meltwater must be at T=0C because it is in contact with the ice. (Because when liquid and solid phases are in contact, the only temperature they can be at is the melting point.)

Therefore, you know that the steam at 100C must become water at 0C, and some of the ice at 0C also must become water at 0C. The steam of course contributes 0.5kg of water, since that's the amount of steam you had to start with. The question is, how much ice melts? The steam must change first from 100C steam to 100C water, then cool to 0C water, while the ice must change from 0C ice to 0C water. The energy released by condensing and cooling the steam is what powers melting the ice, so the energy absorbed by the ice is the same as the energy given off by the steam.

Steam heat loss: 500g*540cal/g + 500g*1cal/gC*100C = 320000cal
Ice hean gain: m*80cal/g.

Since these are equal, m=32000/80 = 4000g = 4kg.

The steam contributed its 0.5kg also, so the total amount of water is 4.5kg.

Hope that helps!

--Stuart Anderson

[caitie16]

that helps alot! thank you so much again for the help (and the quick response), you have a very clear way of explaining things. cheers