Physics Question - Temperature?

Question

A 0.10 kg piece of copper at an initial temp of 95 deg C is dropped into 0.20kg of water contained in a 0.28 kg aluminum caloimeter. The water and calorimeter are initially at 15 deg C. What is the final temp of the system when it reaches equilibrium?
(cp of Copper= 387J/kg x deg C; cp of Aluminum=899J/kg x deg C; cp of Water=4186J/kg x deg C)

I'm totally not getting this!

Answer 1

In the system of copper, water, and aluminum, heat will flow from hot to cold until everything is at the same temperature. At the same time, the heat lost by the copper will equal heat gained by water and aluminum.

So you let T be the final temperature. Then the heat lost by the copper will be dH_cu:

dH_cu=(95-T)*Cp_cu*0.10

The heat gained by aluminum and water is dH_al and dH_h2o, respectively:

dH_al=(T-15)*Cp_al*0.28
dH_h2o=(T-15)*Cp_h2o*0.20

Setting heat lost equal to heat gained:

dH_cu=dH_al+dH_h2o

(95-T)*Cp_cu*0.10=(T-15)*(Cp_al*0.28+Cp_h2o*0.20)

Now solve for T and that's the final equilibrium temperature.

Answer 2

Look at it as a balance,
Heat generated from cooling items = heat uptake from heated items. (heat is conserved)
So for each item, q = w cp del T. Let Tf be the final temp Tf>15
For the copper, 0.1 x 387 x (95-Tf) =
water 0.2 x 4186 x (Tf-15) plus
aluminum 0.28 x 899 x (Tf-15)

You can solve exactly. As a first approximation, thee aluminum shell can be neglected, and you come up with 38.7 (95-Tf) appx 837 (Tf-15). From this, it takes about a 22 degree drop in copper temp to raise the water temp 1 degree. Tf should be around 18 deg C.