Energy Transfer in a Water and Aluminum Pan System

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Homework Statement



100 grams of boiling water (temperature 100° C, heat capacity 4.2 J/gram/K) are poured into an aluminum pan whose mass is 375 grams and initial temperature 24° C (the heat capacity of aluminum is 0.9 J/gram/K).
(a) After a short time, what is the temperature of the water?

(b) Next you place the pan on a hot electric stove. While the stove is heating the pan, you use a beater to stir the water, doing 1400 J of work, and the temperature of the water and pan increases to 74.6° C. How much energy transfer due to a temperature difference was there from the stove into the system consisting of the water plus the pan?


Homework Equations



deltaE=mC*deltaT
deltaE=Q+W

The Attempt at a Solution



Ok I got part A with no trouble but I'm struggling with part B. I tried using the mC*delta T equation for the water and the aluminum pan...adding those values and setting it equal to Q+W. I used 8.6 for delta T (74.6-66) since I got 66 degrees for my answer in part A. I then subtracted W to find Q in the second equation but it's not the correct answer. I think what might be throwing me is that the system consists of the water and pan. Any suggestions?
 
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You didn't say what value of C you used in the second calculation. But C' should be the weighted average of the mass of the water and the aluminium ( British spelling), and M' the sum of the masses.

C' = (C1*M1 + C2*M2)/(M1+M2)
M' = M1 + M2.

I think this will work, but I could be wrong.
 
Mentz114 said:
You didn't say what value of C you used in the second calculation. But C' should be the weighted average of the mass of the water and the aluminium ( British spelling), and M' the sum of the masses.

C' = (C1*M1 + C2*M2)/(M1+M2)
M' = M1 + M2.

I think this will work, but I could be wrong.

I calculated two values of mC*delta T, one for the water (100*4.2*8.6) and one for aluminum (375*0.9*8.6) and added those together and set that equal to Q+1400. I got Q=5115 (same answer as you) and it's incorrect.