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Ice added to water at 273K, calculate new ice mass

  1. Apr 18, 2015 #1
    • Member warned about lack of template
    Hi! I'm new here and I can't find anything helpful through google so I thought I'd give PF a try. Sorry for not using the template fully but I have no ideas on how to solve this problem.

    1. 0.1kg ice at 263K is added to 1kg water at 273K. Calculate new mass of the ice.

    2. I know how to do this calculation if the water temperature is higher than 273K, but then I got this problem where the water is already at 273K (0°C) therefore no change in temperature so I can't use Q=c⋅m⋅ΔT or at least I can't get my brain around how. Is there an "opposite direction" enthalpy formula? As I have understood it this situation will actually increase the mass of the ice since energy will be released when the ice is heated to 273K.


    3. Can I do this the other way around i.e. calculate how much water needs to turn to ice to heat the ice to 273K?

    ΔQ=cice⋅mice⋅ΔT = 2.2kJ/(kg⋅K)⋅0.1kg⋅10K = 2.2kJ
    Then
    2.2kJ = m⋅Lwater
    m = 2.2/332 ≈ 6.6⋅10-3kg


    mice total = mice 0+mice 2 = 0.1+6.6⋅10-3 = 0.1066kg

    Thanks!
     
    Last edited: Apr 18, 2015
  2. jcsd
  3. Apr 18, 2015 #2

    DrClaude

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    Staff: Mentor

  4. Apr 18, 2015 #3
    thank you, I'm uncertain if this is the same we have been thought about enthalpy because in our tables we have these specific formulas for phase changes with lower case l. Like ΔQ=lmm and ΔQ=lvm (m=melt, v=vaporize).

    I've made an attempt at a solution in my OP, can I use this method for any phase change regardless of "direction"?
     
    Last edited: Apr 18, 2015
  5. Apr 18, 2015 #4

    DrClaude

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    I would advise against editing the OP. Things are easier to follow if the development can be seen in the replies.

    Yes. For example, the heat of "melting" can be used both for melting and for solidification.
     
  6. Apr 18, 2015 #5
    Thank you

    Here is my first attempt on a solution, I suspect something is missing.

    ΔQ=cice⋅mice⋅ΔT = 2.2kJ/(kg⋅K)⋅0.1kg⋅10K = 2.2kJ
    Then
    2.2kJ = m⋅Lwater
    m = 2.2/332 ≈ 6.6⋅10-3kg


    mice total = mice 0+mice 2 = 0.1+6.6⋅10-3 = 0.1066kg
     
  7. Apr 18, 2015 #6

    DrClaude

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    That looks fine.
     
  8. Apr 18, 2015 #7
    Woo hoo, thank you! Knowing these properties go any direction clears up a lot of confusion. One final question just to see if I've understood everything correctly. If both ice and water is 273K from the start then nothing will happen?
     
  9. Apr 18, 2015 #8

    DrClaude

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    Staff: Mentor

    Correct. At the temperature of the phase transition, both phases coexist. The amount of one phase or the other can then only change if there is exchange of heat with something outside the system.
     
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