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Final temperature of ice mixed with water

  1. Sep 4, 2015 #1
    What will be the final temperature of 3.00 kg of crushed ice with temperature -3.0 C mixed with 4 liters of water with temperature 10.0 C? We can ignore all the energy exchange with the environment

    What I do know and I do:

    the weight of the ice = 3 kg
    Ice temprature = -3 C
    melting energy = 3 kg * 333 000 J / kg = 999000 J
    weight of water = 4 kg
    specific heat capacity of water = 4.2 Jk / kg K
    specific heat capacity of ice = 2200 Jk / kg K

    I understand just melting energy, but I do not know how will I find out the final temperature. A solution that I have but do not understand why :

    dQ ice = 3 * 2200 * (0 + 3) = 19800 J
    dQ water = 4 * 4200 * 10 = 168000 J
    Melting energy = 999000 J


    End temperature becomes equal to 0. but how can I find it?!
     
    Last edited: Sep 4, 2015
  2. jcsd
  3. Sep 4, 2015 #2
    Break the problem into steps. First heat the ice to 0oC and cool the water to 0oC (as you already have) and then consider the available melting energy. Can you use all of it?
     
  4. Sep 4, 2015 #3

    SteamKing

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    Well, you must do some visualization of what is happening to the ice when it is mixed with water.

    1. The ice cannot melt until its temperature reaches the melting point of ice. Since the ice temperature is initially -3° C, it takes a certain amount of heat to raise its temperature to 0°C. Where does this heat come from?
    2. Once the ice is at 0°C, then it can begin to melt, but it needs even more heat to do this. Where does this heat come from? What happens if there is not sufficient heat to melt all the ice?
     
  5. Sep 4, 2015 #4

    Doc Al

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    Think in terms of steps (You've already done the calculations):

    How much energy is available if the water went to 0 degrees?
    How much energy is required to raise the temp of the ice to 0 degrees?
    How much energy is required to melt all the ice? Is there enough energy?
     
  6. Sep 4, 2015 #5
    Ok and thank you guys. I think I understand that Ice will reach 0 from -3 and water will do that too from 10 to 0. For Ice we have a melting energy and (Q ice). For water (Q water) but if this is so why should we calculate energies?!
     
    Last edited: Sep 4, 2015
  7. Sep 4, 2015 #6
    1- We need a melting energy which is 999000 J. I think the ice raises 0 with 999000 J and becomes liquid with dQ ice = 19800 J
    2- In this part the ice is mixed with water which is 10 C. I think dQ water = 168000 J helps the water and ice to reach 0.
    But I am on the right path?
     
  8. Sep 4, 2015 #7

    Doc Al

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    You need the energy contained in the water to see if it is enough to (1) raise the temp of the ice to the melting point, and (2) melt all the ice. If there's energy left over, that means the newly formed water will end up with a final temp above zero.
     
  9. Sep 4, 2015 #8
    OK I see that dQ water = 168000 J, dQ ice=19800 J and melting energy of ice = 998000

    dQ ice + melting energy of ice > dQ water

    or dQ water / (dQ ice + melting energy) is less than 1.


    And because of this we can say the final temperature is 0 ?!
     
  10. Sep 4, 2015 #9

    Doc Al

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    Sure.

    Think in steps:
    Can you raise the temp of the ice to freezing? Yes! How much of the water's energy is left?
    Is there enough to melt all the ice? No! So at some point, the water will get to 0 degrees before all the ice is melted. Final temp must be zero.
     
  11. Sep 4, 2015 #10
    Thank you so much for help.
    At first time, I was just thinking like this

    dQ loss = dQ gain
    or
    dQ ice = dQ water

    But sometimes I confuse these types of issues with other calorimetry calculation
     
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