# Entropy-any direction would be appreciated

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

## The Attempt at a Solution

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Andrew Mason
Homework Helper
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
This does not involve an entropy calculation. This is simply a heat flow calculation. You have to know the specific heat of Al, Water and Ice as well as the heat of fusion of water/ice. Then you have to set up equations for the change in temperature in terms of the heat flow from the water/cup to the ice and the final temp. Then solve the equations.

You can't convert Joules/g C to Calories. You can convert Joules/g C to Calories/g C by dividing by 4.187 Joules (= 1 Cal.)

AM

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