Mix steam with colder water, what is final temp?

[SunshineCat]

Homework Statement

100 g of steam at 100C is mixed with 1 kg of water at 10C. What is the final temperature of the mixture? (in C)
heat capacity of water: cw = 4.186 kJ kg-1 K−1
latent heat of vapourisation for water: Lv = 2256 kJ kg−1
latent heat of fusion for water: Lf = 334 kJ kg−1

Homework Equations

Q=mL
Q=mcΔT

The Attempt at a Solution

I think that there needs to be a phase change of condensation, 0.1kg of 100C steam to 0.1kg of 100C water, so this will be Q=mL and will be an energy loss
The energy lost will go to the colder water
After this, energy will keep flowing from the 100C water to 10C water until they reach the same temperature, energy gain and loss will be Q=mcΔT
I can't seem to get the calculation right, I keep getting 60C when the answer is 67C

ΔQ (steam) = ΔQ(water)
Q=mcΔT = mL + mcΔT
0.1x4.186x(T-100) = 1x4.186x(T-10) - 0.1x225.6
225.6=3.77T
T=60C?

 

[CWatters]

0.1x4.186x(T-100) = 1x4.186x(T-10) - 0.1x225.6

That should be 0.1 * 2256

 

[SunshineCat]

That should be 0.1 * 2256

Oh yeah, that was a typo I think