Physics Calorimetry problem?

M

Michelle S

New member
0.25 kg of water at 90 degrees C is added to 0.35 kg of water at -10 degrees C in an aluminum cylinder of mass 0.100 kg also at -10 degrees C. What is the final temperature? Water's specific heat is 4186 J/kgK. Aluminum's specific heat is 9.00x10^2 J/kgK

can someone show me how to this step by step?
 
You said water at -10 C and didn't give the heat of fusion of water, so I'm assuming it's supercooled water!
Tf = 30.225 C
Step by step:
Basic equation for final (f) state of an N-component mixture expresses the equality of initial and final heats: mfcfTf = Σ(i=1,N)(miciTi)
Since mfcf = Σ(i=1,N)(mici),Σ(i=1,N)(mici)Tf = Σ(i=1,N)(miciTi). This leads to
Tf = (Σ(i=1,N)(miciTi))/Σ(i=1,N)(mici) = 30.225 C
 
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