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Latent Heat

  1. May 4, 2008 #1
    1. The problem statement, all variables and given/known data
    In an insulated vessel, 250 g of ice at 0 °C is added to 600 g of water at 18 °C. (a) What is the final temperature of the system? (b) How much ice remains when the system reaches equilibrium?

    2. Relevant equations

    Q = mL, Q = mc[tex]\Delta[/tex]T

    3. The attempt at a solution
    If I'm not mistaken: heat gained by ice = - heat lost by water

    [tex]Q_{ice} = -Q_{water} [/tex]

    [tex] \Rightarrow m_{ice}L_{ice} + m_{ice}c_{water} \Delta T = - m_{water}c_{water} \Delta T [/tex]

    [tex] \Rightarrow (0.25 kg)(3.33* \times 10^5 J/kg) + (0.25 kg)(4186 J/kg \cdot K)(T_{f} - 273.15 K) =
    - (0.6kg)(4186 J/kg \cdot K)(T_{f} - 291.15 K) [/tex]

    Which when I solve for [tex]T_{f}[/tex] gives me a ridiculous number.

    I've looked at the answer and it's 0.0 °C. What am I doing wrong?
  2. jcsd
  3. May 4, 2008 #2


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    Homework Helper

    Hi caesius,

    It appears not all of the ice has melted; so the heat absorbed by the melting ice is not


    since only part of that mass has melted. How can you find out how much of the ice has melted?
  4. May 4, 2008 #3
    If I introduce an unknown mass of ice will I not have too many unknowns? I'm still stuck.
  5. May 4, 2008 #4


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    Homework Helper

    When you have a problem like this, probably the first thing to find out is if the ice completely melts or not. You can find this by calculating two separate things:

    total amount of heat loss to bring the water 18 degrees to zero degrees

    total amount of heat required to completely melt 250 grams of ice

    What do you get for these two numbers? Is there enough heat lost in the water going from 18 to 0 degrees to completely melt the ice?

    Once you have this information, that will tell you how to set up the general equation of heat lost=heat gained.
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