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Entropy-any direction would be appreciated

  1. Jan 31, 2008 #1
    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. 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
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 31, 2008 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    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
     
    Last edited: Jan 31, 2008
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