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Determining latent heat of fusion of ice

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data
    For the problem below I was able to work out part (i) but I am stuck on working out part (ii) I tried using information given but was unsuccessful. Can anyone help me out?

    An insulated copper container of mass 0.250 kg contains 0.350 kg water. Both the container and the water are at 25.0 degrees. 0.012 kg ice at 0.0 degrees is added to the container. Eventually the container and contents reach thermal equilibrium at 21.7 degrees.

    We know:
    Specific Heat of water = 4186 J/Kg Degree
    Specific Heat of copper = 387 J/Kg Degree
    Latent heat of vaporisation of water = 2.26x10^6 J/Kg

    (i) What is the total heat released (in J) by the copper container and the 0.350 kg water as they cool down from 25.0 degress to 21.7 degrees?

    (ii) Determine the latent heat of fusion of ice.



    2. Relevant equations
    Q=mc[itex]\Delta[/itex]T
    Q=mL

    3. The attempt at a solution
    (i) What is the total heat released (in J) by the copper container and the 0.350 kg water as they cool down from 25.0 degress to 21.7 degrees?
    Qw=0.350x4186x(25-21.7) = 4835J
    Qcopper = 0.250x387x(25-21.7) = 319J

    QT=Qw+Qc = 4835+319 = 5154J
     
  2. jcsd
  3. Nov 14, 2011 #2
    The container with the water in it has cooled down. The energy emitted went into the ...
     
  4. Nov 14, 2011 #3
    I don't know why the heat of vaporization is mentioned. Nothing is boiling.

    Think about what happens when the ice melts. It absorbs heat when it does. This occurs at constant temperature. So now you have liquid water but it is at 0 C. More energy is used to get it to the final temperature. Put what I've typed into an equation.
     
  5. Nov 14, 2011 #4
    You have been given the wrong value !!!For freezing/melting you should have been given the specific latent heat of fusion (330,000J/kg).
    You have been given the specific latent heat of vaporisation which relates to evaporation/condensation
     
  6. Nov 14, 2011 #5
    correction!! You have to calculate the latent heat of vaporisation.... you are correct, you do not need the latent heat of vaporisation
     
  7. Nov 14, 2011 #6
    I agree with your value of QT. This is the heat energy given up by the container and the water. This heat did 2 jobs
    1)some heat needed to convert the ice at 0C into water at 0C (latent heat)
    2)some heat to raise the temp of the melted ice from 0C to 21.7C
    I hope you can put the appropriate numbers in here to find the latent heat of ice ( I got 337,000 J/kg)
     
  8. Nov 14, 2011 #7

    sophiecentaur

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    We used to do loads of these problems when I was at School (instead of being told about Relativity and Quantum - it was a long time ago!!!)
    It was called the Method of Mixtures.

    Work out the Heat energy at the start (temperature can be relative to 0C or anything else you might choose) then equate it to the Heat energy at the end. This is more or less what you did but you need the total heat at the end ('sensible heat' plus latent heat) and equate the two. The final 'sensible heat' needs to involve the eventual temperature of the copper, original water and the melted ice.

    The term 'sensible heat' is an old fashioned steam engineering one but is quite good because it is the heat you could 'sense' by touching (i.e. it involves the temperature)
     
  9. Nov 14, 2011 #8
    You have got practically the correct answer, the rounding is a bit different but that is no big deal, the method to get there is, the answer was 3.39 x 10^5 J/Kg.

    I understand the formula to use, L=Q/m, however I am not sure which values I need to find Q for this particular section. I have the right idea I just can't quite find the right direction at the moment.

    And Sophie, they call it specific heat now instead of sensible heat, well atleast where I study :smile:
     
  10. Nov 14, 2011 #9

    sophiecentaur

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    @savva
    No. 'Specific Heat' is the same defined term that has always been used. Sensible Heat is the Mass times specific Heat times the temperature rise. (Or Heat Capacity times temperature rise). It's not used much but is actually describes, in two words, what would take more words, expressed in another way.
    The third 'type' of heat in steam engineering is Super Heat (what you get when you have steam above 100C and under pressure).
    The word 'Specific' occurs all over and, I think, usually refers to Intrinsic Quantities (i.e. independent of the quantity of material involved. 'Specific Gravity', for instance.
     
  11. Nov 14, 2011 #10
    Sorry, you are right specific heat = c, in our text book they don't have a name for Q (sensible heat) - they just mention that it is the energy required to change the temperature of an object.
     
  12. Nov 14, 2011 #11

    sophiecentaur

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    It's a very quaint term. No wonder they dropped it.
     
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