If 281 g of water is contained in a 346 g?

Question

If 281 g of water is contained in a 346 g
aluminum vessel at 6 C and an additional 85 g
of water at 81 C is poured into the container,
what is the final equilibrium temperature of
the mixture? Assume the specific heat of aluminum is 900 J/kg C. The specific heat of water is 4186 J/kg
C. Answer in units of C.

Answer 1

Since no heat is added or escapes, this is a simple matter of moving it around.

The heat gained by the aluminum vessel and its contents must match the heat lost by the added water.

Let Tf be the final temperature. Raising the temperature of the aluminum cup requires energy equal to:

Qa = Ca M dT where:
Ca is the specific heat of aluminum
M is mass of the aluminum cup
dT is the difference between the initial and final temperatures

http://en.wikipedia.org/wiki/Specific_heat_capacity

Similarly, Qc and Qw for the energy taken by the initial contents of the aluminum vessel and the energy given by the added water.

All three are a function of the final temperature and we know from conservation of energy that:

Qa + Qc = Qw

so you just solve for the Tf that makes this equation true.