Calculating Final Temperature of Steam and Water Mixture - Chemistry Question

[GodsChild086]

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]

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]

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]

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]

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...