Physics calorimeter question

Question

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

Here is my problem: I know how to set up the problem but am making algebra mistakes. I do not understand why Q= 0 the Q of the system

the equation sets up as (.2kg)(4.19 X 10^3 J/kgC) (Tf-10 C) (.3 kg)(900 J/kgC)(Tf- 10 C) (.1kg)(4.19 x 10^3 J/kgC)(Tf- 100 C) = 0

question is how is the math done algebraic wise (factoring out the Tf) and why is it = 0????? Please help anyone!

Answer 1

the equation is Q(hot) + Q(cold) = 0

it is 0 because the heat lost from the hotter object goes into raising the temperature of the cooler object.

0.2(4.19x10^3)(Tf-10) + 0.3(900)(Tf-10) = 0.1(4.19x10^3)(Tf-100)