What Will Be the Final Temperature of Water After Adding Steam?

[Jm4872]

Homework Statement

What will be the final temperature of the water in an insulated container as the result of passing 5.00 of steam at 100.0 into 100.0 of water at 27.0?

Homework Equations

q=mct
q=nH

The Attempt at a Solution

I am completely lost as to how to solve for final temperature, I attempted it, but it didn't work, this is what I did.

n=5.00g/(18.02g/mol)=0.277 mol

q=(0.277 mol)*(40.6KJ/mol)=11.27KJ
11.27KJ=mct
11.27KJ=(100g)*(4.18)*(Tf-27)
(11.27/418)=Tf-27
Tf=(0.02696)+27=27.0

can someone please help, I am so lost :( I don't understand what it is that I'm doing wrong

 

[Borek]

passing 5.00 of steam at 100.0 into 100.0 of water at 27.0?

5 of what, 100 of what, 100 of what, 27 of what?

 

[Jm4872]

5 of what, 100 of what, 100 of what, 27 of what?

oops, I didn't realize that the values didn't show up.

it should be 5.00g, 100degrees celcius, 100.0g and 27 degrees celcius

 

[Borek]

11.27KJ=(100g)*(4.18)*(Tf-27)

4.18 of what?