This is a really simple problem, but I just can't remember how to combine the container and the water. 1. The problem statement, all variables and given/known data An aluminum container weighing 200g contains 500g of water at 20*c. A 300g shot of the same aluminum at 100*c is put into the water. The specific heat for the aluminum is 0.215cal/(g*c). What is the final temperature for the entire system? 2. Relevant equations dQ = c*m*dT 3. The attempt at a solution (T - 20*c) x [(500g x 1cal/(g*c)) + (200 x 0.215cal/(g*c))] = (T - 100*c) x 300g x 0.215cal/(g*c) T = 9.1216*c However this answer seems really off.