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How much ice melts in water?

  1. Aug 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Ice at 0°C is added to a 1 kg of lemonade cooling it from 20°C to 0°C. How much ice melts? (The heat capacity of lemonade equals that of water.)


    2. Relevant equations
    Q = mcΔT


    3. The attempt at a solution
    Found the heat in of the lemonade to the ice... = 83.38 kJ and don't know where to go from there. c of water is 4.169.
     
  2. jcsd
  3. Aug 13, 2014 #2

    gneill

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    Staff: Mentor

    What happens to the ice that melts? What's involved?
     
  4. Aug 13, 2014 #3
    A phase change occurs, but the problem does not give the mass of the ice.
     
  5. Aug 13, 2014 #4

    gneill

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    Staff: Mentor

    Right. Because it's the mass of ice that melts that you're looking for.... Write out the formula that pertains to that phase change. Which of the variables do you you have values for?
     
  6. Aug 13, 2014 #5
    How much heat is removed from from the lemonade when it cooled from 20C to 0 C?

    Chet
     
  7. Aug 13, 2014 #6
    gneill: Ok, so the formula that pertains to that phase is ΔQ = Lm. I know that L is the latent heat of fusion for water so it is 333 kJ/kg. Would the change in heat be the 3.38 kJ?

    Chestermiller: I found the change in heat for the lemonade form 20-0 °C to be 83.38 kJ.
     
  8. Aug 13, 2014 #7
    Sorry, i meant the 83.38 kJ
     
  9. Aug 13, 2014 #8

    gneill

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    Staff: Mentor

    Yes. You calculated the heat that needed to be extracted from the lemonade in order to cool it to 0C, and that heat had to go somewhere. Clearly it went to melt some of the ice, providing the heat of fusion for the process.
     
  10. Aug 13, 2014 #9
    So if the heat removed from the lemonade was 83.38 kJ, and the amount of heat required to melt 1 kg of ice is 333 kJ/kg, how many kg of ice have to melt?

    Chet
     
  11. Aug 13, 2014 #10
    Right, so I got 83.38 kJ/(333 kJl/kg) = .25 kg. So .25 kg. Is this right?
     
  12. Aug 13, 2014 #11
  13. Aug 13, 2014 #12
    Great! Thank you Chestermiller and gneill!
     
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