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Latent Heat Question

  1. Mar 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A well insulated cup contains 125 g of water at 23 degrees. How many grams of ice must be added to lower temp to 12?

    2. Relevant equations

    Q = c*m*delta t

    3. The attempt at a solution

    energy needed to lower temp:
    Q = 4190 J / kg * (-11 degrees) * (.125 kg) = -5760 J

    Q from ice = - Q from water = 5760 J

    to find mass of ice needed:

    5760 J = Latent heat of ice * mass needed = (3.34 * 10^5 J / kg) * m

    = .0172 kg = 17.2 g

    The correct answer is 15 g. What am I doing wrong?
     
  2. jcsd
  3. Mar 11, 2009 #2

    tiny-tim

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    Hi veronicak5678! :smile:
    hmm … you didn't stir! :wink:
     
  4. Mar 11, 2009 #3
    I'm sorry, but I don't follow...
    Should I be using something other than latent heat?
     
  5. Mar 11, 2009 #4

    tiny-tim

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    yes … the ice will become water, and mix with the water already there :smile:
     
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