How Does Adding an Aluminum Calorimeter Affect Heat Transfer Calculations?

[ariana0923]

Homework Statement

A 0.10 kg piece of copper at an initial temperature of 95°C is dropped into 0.20 kg of water contained in a 0.28 kg aluminum calorimeter. The water and calorimeter are initially at 15°C. What is the final temperature of the system when it reaches equilibrium?
M(w) = 0.20kg
Cp,(w) = 4186J/kg°C
M(Al) = 0.28kg
Cp,(Al) = 899J/kg°C
T(w) = T(Al) = 15°C
M(Cu) = 0.10kg
Cp,(Cu) = 387J/kg°C
T(Cu) = 95°C


I'm confused because usually there is just the water and the substance, but now there's the aluminum calorimeter to account for. I'm not sure what to do..do I still use the Qw= -Qx approach? If so, do I add the aluminum info to the water info...someone else seemed to do it that way, but he and i got answers differing by about 1 degree.

Homework Equations

-Q(Cu) = Q(Al+w)

The Attempt at a Solution

[cp(cu)] [M(cu)] [ΔT(cu)] = [cp(al) + cp(w)] [M(al) + M(w)] [ΔT(al and water)]
(-387)(0.10)(Tf-95) = (899+4186)(0.28+0.20)(Tf-15)
Tf=16.77 degrees C

**Before, I made the right side negative (Q al+W) and got 16.25, almost the same thing, so I don't know if that matters, or if I did the math wrong. Anyhow, is my above work correct?

 

[JazzFusion]

..do I still use the Qw= -Qx approach?

YES!

If so, do I add the aluminum info to the water info...? ...

-Q(Cu) = Q(Al+w)

[cp(cu)] [M(cu)] [ΔT(cu)] = [cp(al) + cp(w)] [M(al) + M(w)] [ΔT(al and water)]

Not quite. Instead, use a different term for the contribution of each. Try this:

-Q(Cu) = Q(Al) + Q(w)

-[cp(cu)] [M(cu)] [ΔT(cu)] = cp(al)M(al)[ΔT(al)] + cp(w)M(w)[ΔT(w)]

The water will gain heat energy. The aluminum cup will gain heat energy. Both will start at the same T (15 degrees, in this case), and both will end at the same temperature (the equilibrium temperature you are trying to find). In other words, ΔT(w) = ΔT(al).