# Specific Heat

1. Nov 25, 2008

### XenoPhex

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.

2. Nov 25, 2008

### mgb_phys

Energy is conserved so, energy gained by:
Econtainer = mct = 200 * 0.215 *(T-20)
Ewater = mct = 500 * 1* (T-20)
Equals energy lost by:
Eshot = mct = 300 * 0.215* (100-T)

(Final temperature T)

3. Nov 25, 2008

### XenoPhex

Wait, so would you just make Eshot = Ewater and then solve for T and then Eshot = Ewater and solve for a different T?

4. Nov 25, 2008

### mgb_phys

You assume everything comes to the same final temperature. Eshot = Ewater+Econtainer
Watch the signs, energy is lost by the shot and gained by the container and water.