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Latent heat

  1. Jul 31, 2009 #1
    1. The problem statement, all variables and given/known data
    You are given 2.5 102 g of coffee (same specific heat as water) at 70.0°C. In order to cool this to 60.0°C, how much ice (at 0.0°C) must be added? Neglect heat content of the cup and heat exchanges with the surroundings.


    2. Relevant equations
    Q=(m ice)(c ice)(delta T ice+ m ice)(Lf)
    Q=(m coffee)(c coffee)(delta T coffee)


    3. The attempt at a solution
    If these are the right formulas I still have the problem that there are 2 variables, since I don't know Q or m ice.
     
  2. jcsd
  3. Jul 31, 2009 #2

    mgb_phys

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    The Q's in both equations are the same, energy lost by coffee = energy gained by ice.
    so you simply set the two equations equal to eahc other
     
  4. Jul 31, 2009 #3
    do you mean (m ice)(c ice)(delta T ice+ m ice)(Lf)=(m coffee)(c coffee)(delta T coffee)?
     
  5. Jul 31, 2009 #4

    mgb_phys

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    Almost, remember ther is energy to melt the ice (which doesn't depend on temperature) and then the energy to heat the resulting water to the final temperature.
     
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