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What is the final temperature?

  1. Jul 18, 2013 #1
    30g of ice at -15 degree celcius is added to 100g of water at 25 degree celcius. What is the final temperature?

    T1=ice
    T2=H2O

    at equilibrium must have
    m1c1(Te-T1)+m1k=m2c2(T2-Te) [k=3.3x10^5 J/K fusion heat coeff.]

    then
    Te=(m1c1T1+m2c2T2-m1k)/(m1c1+m2c2)
    T1=268 k
    T2=298 K
    c1=2090 J/(kg K)
    c2=4186 J/(kg K)
    Te=(0.02*2090*268+0.1*4186*298-0.02*3.3… / (0.02*2090+0.1*4186)
    Te=280,9 K
    Te=7.8 C

    Final Temperature is 7.8 C

    does this look ok?
     
  2. jcsd
  3. Jul 18, 2013 #2

    PeterO

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    Can't quite follow your calculations but looking at it conceptually.

    Specific heat for ice is about half that of water, so you will need much less heat to warm the ice up to zero degrees than it takes to cool the water to zero degrees.
    Less mass of ice, smaller temperature change and easier to change so a win on all three counts.

    Big question - how much heat do you need to melt the ice?

    If you need more heat to melt the ice than you need to cool the water to zero degrees you will end up with an ice/water mix at zero degrees.
     
  4. Jul 18, 2013 #3
    hmm.. my tutor help me to solve this problem but now I look at it again, some of these equation don't make sense. any suggestion? :(
     
  5. Jul 18, 2013 #4

    SteamKing

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    How many kg is 30 g of water?
     
  6. Jul 18, 2013 #5

    SteamKing

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    Rather than write one huge equation which no one can understand, and which is prone to error, look at the problem in stages:

    1. Find the amount of heat transfer to the ice to bring its temp. to 0 deg. C
    2. Find the amount of heat transfer to the ice to cause its phase change from solid to liquid.
    3. Find the equilibrium temperature of the now melted ice and the cooled water.

    In this way, it is easier to detect and fix any mistakes in the calculations.
     
  7. Jul 18, 2013 #6

    CWatters

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    There doesn't appear to be enough energy in the starting components for all 130g to be water.

    One way to do this is to work out the energy contained in the starting components (eg relative to absolute zero). Then apply conservation of energy... pretend you have 130g of ice at abs zero and see how hot it would end up be if you added that energy.

    There appears to be enough to raise it all to 273K (aka 0C) but not quite enough left to melt it all to water.
     
    Last edited: Jul 18, 2013
  8. Jul 18, 2013 #7
    how about this one?
    latent heat of fusion(L) = 80 cal g
    Mc = mass of ice

    heat gained = (Mc *4180 * 15) + (Mc * L) + (Mc * 4180 * t)
    heat gained = (30 * 4180 *15) + (30 * 80) + (30 * 4180 * t)
    t = -15c

    final temp is -15c
     
    Last edited: Jul 18, 2013
  9. Jul 18, 2013 #8

    CWatters

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  10. Jul 18, 2013 #9

    PeterO

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    If you are going to use 80 for the latent heat, you need to use 1 for the SHC of water and about 0.5 for the ice.*

    To heat the ice to 0 degrees you need

    0.03 x 15 x 2090 Joules.

    To melt all that ice you need 0.03 x 3.3 x 105 J

    So to get the ice from -15 up to water at 0 degrees you need the sum of those two.

    To Cool the water to 0 degrees you need to extract

    0.1 x 25 x 4180 Joules

    That is more than enough to get the ice up to zero degrees, but nowhere near enough to melt all the ice - so you will be left with a mix of ice and water - naturally at zero degrees.
     
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