What is the Final Temperature When Ice is Added to Water in an Aluminum Cup?

[hils0005]

A cup made of 100g of aluminum holds 200g of water at 50deg C, a 10g piece of ice is placed in the cup of water, what is the final temperature?

2. Homework Equations
Q=L(f)m
S=cm lnTf/Ti
specific heat of aluminum 0.90 J/gC

The Attempt at a Solution

I think the first step is to determine S(ice)
Q=Lm=79.5 cal/g * 10g=795cal
S(ice)=Q/T=795/273K=2.91
S(water)=Q/T=-795/323K=-2.46
2.91-2.46=.45cal/K

then as 0deg C water goes to 50deg C?
I believe I should be using an integral as the ice will be changing the temp of the water?
S(ice)=cmln(Tf/Ti)
1cal/gC * 10g ln(795/323)=9

S(water)=cmln(Tf/Ti)
1cal/gC*200 ln(795/323)=180

So now I'm confused-because I believe I would need to know the final temp of the water before calculating S with the Steel??
I also do not understand how to convert J/gC to cal

 

[Greg Bernhardt]

The final temperature in physics is absolute zero. The coldest temperature known. It is impossible to go below this and remain in our normal physical form. To do so, we would cease to exist as we know ourselves and become a different kind of matter. This is why we can’t go to absolute zero. We don’t have the technology to do so.