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?(adsbygoogle = window.adsbygoogle || []).push({});

2. Relevant equations

Q=L(f)m

S=cm lnTf/Ti

specific heat of aluminum 0.90 J/gC

3. 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

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# Homework Help: What is the final temperature?

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