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Heat problem

  1. Nov 14, 2003 #1
    The question is "8 grams of water at 100 degrees C are poured into a cavity in a very large block of ice initially at 0 degrees C. How many g of ice melt before thermal equilibrium is attained round off to the nearest whole number?"

    As I understnd it I need to:
    (heat to change ice to water)+(heat to raise water from 0degrees C to T)=(heat lost by 8g of water cooling from 100degrees C to T)

    My book shows the formula for this is:
    (mass of the ice*Lf) + (mass of ice*c)=(mass of water*c)(100degrees-T)

    So:
    m(333kJ/kg)+m(4186J/kg*degC)*(T)=.008kg(4186J/kg*degC)*(100degC-T)

    The m is what I'm solving for, what do I do about the T? I know its in equilibrium, but I don't know what it is. I can't solve for both.

    Thanks in advance!
     
  2. jcsd
  3. Nov 14, 2003 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    "8 grams of water at 100 degrees C are poured into a cavity in a very large block of ice initially at 0 degrees C."

    A "very large block of ice". That means that there is sufficient ice that some will still be frozen when equilibrium is reached. The equilibrium temperature is still 0 degrees C. Take T= 0.
     
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