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GodsChild086
Nov24-04, 08:29 PM
I'm really stuck on this question:

What will be the final temperature if 50.0 g of steam at 100 degrees celsius is mixed with 600 mL of water at 10.0 degrees celsius? (Answer: 58.3 degrees celsius)

So far I've got:

heat lost (steam) = heat gained (water)
nH + mc(delta)t = mc(delta)t

Sorry I don't know how to make the delta symbol.

(500g/18.02g/mol)(40.65 kJ/mol) + 0.500 kg(2.02 kJ/kg*C)(100-t) = 0.600 kg(4.19 kJ/kg*C)(t-10)

When I calculated everything, I didnt' get 58.3*C, so maybe someone can tell me where I went wrong in either my formulas or calculations. Thanks.

Sirus
Nov24-04, 08:41 PM
You used the specific heat capacity of steam, but after the phase change, it is water that is being cooled and no longer steam.

primarygun
Nov25-04, 03:42 AM
Try to assume that all the steam became liquid water after mixing with water.
If the value of temperature is above 100, then there are some steams left.

GodsChild086
Nov25-04, 11:42 AM
You used the specific heat capacity of steam, but after the phase change, it is water that is being cooled and no longer steam.

I got the answer now....I forgot that after when steam goes through a phase change, it is now water, not steam. Now my answer matches in one in the book. Thank you!

GodsChild086
Nov25-04, 11:48 AM
This is the question that I'm actually really having trouble cause I don't have a clue where to start. I think this is one of those really nasty questions.

Calculate the final temperature when 100 g of ice at -10.0*C are added to 30.0 g of steam at 110*C.

I actually sort of know where to start....but yeah, I'm very stuck on this one...