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).

 

[tomcue]

I cannot confirm the accuracy of your answer without verifying your calculations. However, your approach seems to be correct. You correctly used the principle of energy conservation, which states that the energy lost by one component (in this case, the copper) must be equal to the energy gained by the other components (the aluminum and water).

As for the discrepancy in your answer compared to someone else's, it could be due to rounding errors or differences in the values used for specific heat capacities. It is important to use accurate and consistent values in order to get an accurate result. Additionally, it is always a good idea to double check your calculations to ensure accuracy.

Overall, your approach to solving this problem is correct, but it is important to also pay attention to the details and be careful with your calculations in order to get an accurate result. Keep up the good work!