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Latent heat

  1. Apr 26, 2012 #1
    1. The problem statement, all variables and given/known data

    An aluminum container whose mass is 205g contains 300g of water at 20°C. In this container is then placed 250g of iron at 150°C and 20g of ice at -10°C. Find the final temperature of the mixture.

    Given:

    mAl=205g
    mH2O=300g
    TAl1=TH2O1=20°C
    mFe=250g
    mice=20g
    TFe=150°C
    Tice=-10°C

    2. Relevant equations

    Since the heat gained by aluminum container, water, and ice is equal to heat lost by iron, then

    QAl+QH2O+Qice=QFe

    Q=mcΔT

    3. The attempt at a solution

    What is asked is the final temperature of mixture, Tmix. so,

    QAl=(mAl)(cAl)(Tmix-TAl1)=(205g)(.21cal/g°C)(Tmix-20°C)=43.05Tmix-861

    QH2O=(mH2O)(cH2O)(Tmix-TH2O1)=(300g)(1cal/g°C)(Tmix-20°C)=300Tmix-6000

    Qice=(mice)(cice)(Tmix-Tice)+(mice)(Lf)=(20g)(.5cal/g°C)(Tmix-(-10°C))+(20g)(80cal/g)=10Tmix+100cal+1600cal

    Lf is the latent heat of ice


    QFe=(mFe)(cFe)(TFe-Tmix)=(250g)(.11cal/g°C)(150°C-Tmix)=4125cal-27.5Tmix

    Therefore,

    (43.05Tmix-861cal)+(300Tmix-6000cal)+(10Tmix+100cal+1600cal)=4125cal-27.5Tmix

    380.55Tmix=7564cal

    Tmix=19.88°C

    Is this correct? i think there's something wrong in my solution..
     
  2. jcsd
  3. Apr 26, 2012 #2
    Do this simple check to determine yourself whether your answer is correct. Take your final answer of 19.9 C and look at energy balances. You have energy lost by the vessel and its water. It is lost because the final temperature is lower than its initial temperature. Now determine how much energy the ice gained going from -10 C to 19.9 C including latent heat. Do the same for the energy loss of the iron going from 150 C to 19.9 C. Do you have a balance? If not, you did something wrong.
     
  4. Apr 26, 2012 #3

    i already check it. and yes it's wrong.

    is my assumption correct?
     
  5. Apr 26, 2012 #4
    Yes, the heat gained by the aluminum container, water, and ice equals the heat lost by iron.

    Your answer is off by about 4.5 C. I have not gone through your calculations; I only did my own.
     
  6. Apr 26, 2012 #5
    thank you very much sir!! i finally got the correct answer. i made a wrong computation on the right hand side of the last equation. God bless!
     
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