If 200 g of water is contained in a 300 g aluminum vessel at 10 C and an additio

Question

If 200 g of water is contained in a 300 g aluminum vessel at 10 C and an additional 100g of water at 100 C is poured into the container, what is the final equilibrium temperature of the mixture?

I think I made a math error, but can't find it.

delta Q= (.2kg)(4.19 x10^3 J/kg*C)(Tf- 10C) + (.3) (900J/kg*C)((Tf- 10C) + (0.1 kg)(4.19 X 10^3 J/Kg*C)(Tf-100C)= 0

delta Q (838Tf - 8380) + (270Tf - 2700) + (4197Tff - 4190)
delta Q = 1527Tf - 15270
15270 = 1527Tf
Tf = 10 Which is NOT the correct answer, it should be 35 C.
Thanks!

Explanation / Answer

(0.1 kg)(4.19 X 10^3 J/Kg*C)(Tf-100C)=419Tf-41900 (you did this calculation wrong)
Now, solving,
1527Tf=52980
Tf=34.7 C

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