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What is the final temperature of the contents of the container?

  1. Aug 12, 2005 #1
    A 50 gram piece of ice at -20 C is dropped into an insulated container holding 250 grams of water at 25 C. What is the final temperature of the contents of the container?

    I can't seem to get this problem. I was doing the following:

    Lf water = 333 J/g
    c water = 4186 J/kg*C
    c ice at -10 C = 2220 J/kg C

    .05 kg*2220 J/kg*C * (-10 + 20) (should I use this step?) + .05 kg*4186 J/kg*C * Tf + 10) + 50 g*333 J/g = - .25 kg*4186 J/kg C * (Tf - 25)

    And I keep getting it wrong. Any suggestions? Thanks a ton
     
  2. jcsd
  3. Aug 14, 2005 #2
    Anyone? thanks
     
  4. Aug 14, 2005 #3

    brewnog

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    Units of temperature?
     
  5. Aug 14, 2005 #4

    HallsofIvy

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    To start with, I'm going to assume that the ice will melt.

    First, how much heat is required to bring the ice up to 0 degrees C? Giving up that much heat will reduce the temperature of the water to what? Now, how much heat does it take to melt the ice? Giving up that much heat will reduce the temperature of the water to what? (If 0 or below you're done- there will be a mixture of ice and water at 0 degrees.) Finally, determine the temperature that will give you an equilibrium for both the orginal water and the water from the ice.

    IF giving up enough heat to bring the ice up to 0 degrees would reduce the temperature of the water below 0, the ice will not all melt. You would need to calculate how much heat will be given up when the water freezes and then find the equilibrium temperature (below zero) for the ice. Again, if the heat given up when the water freezes is more than enough to reduce the temperature of the ice to 0, you will have a mixture of water and ice at 0 degrees.
     
  6. Aug 14, 2005 #5
    Thank you very much. I think I followed what you said pretty well, but for some reason my answer is about 1 or 2 C off. What did I do wrong? Thanks a lot.

    (2220 J/kgC*0.05 kg*(0 + 20 C)) = -(4186 J/kgC*0.25 kg*(Tf - 25 C))
    Tf = 24.03 C

    (4186 J/kgC*0.05*(Tf - 0)) + (333 J/g*50 g) = -(4186 J/kgC*0.25 kg*(Tf - 24.03 C))
    Tf = 6.767 C

    The answer is actually 5.81 C, I can't tell what I've done wrong! Thanks a lot for any help...
     
  7. Aug 15, 2005 #6
    I don't mean to keep begging, but if anyone could take a look and tell me if they see anything I'm missing i am really appreciative. My final is tomorrow.
     
  8. Aug 15, 2005 #7
    The LHS of the equation should include sensible heat gain by ice from -20C to 0C, (Don't get misguided by the specific heat at -10C. This is to give you an average specific heat over a range of -20 to 0C.) latent heat gain from 0C ice to 0C water and finally sensible heat gain from 0C water to a final temperature Tf. The RHS is the heat lost by water from 25C to Tf. Apply this and you will get your answer.
     
  9. Aug 15, 2005 #8
    Thank you! I don't see why the way I was doing it doesn't produce the same answer, but it's definitely working right now. Thanks a ton.
     
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