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Saline and Fresh Water Questions

  1. Mar 30, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider two otherwise identical columns of water: one fresh, the other with an initial salinity of 35 pus. Both are initially isothermal at 4°C and each column is 100m deep. There is no horizontal mixing. Both columns are then cooled slowly under calm conditions until ice forms to a thickness of 1m.

    2. Relevant equations

    I've managed most my self but am stuck on:

    (e) Once the ice has formed, how has the depth of each column (from the bottom of the column to the top of the ice) changed?

    (f) One the ice has formed, how has the pressure at the bottom of each column changed?

    3. The attempt at a solution

    My idea was, using ρgh, one would could determine the pressure exerted by the ice cap.

    Taking ρ = 916.7 kg / m^3, I get 1 * 916.7 * 9.80665 = 8,989.8 Pa.

    Because the water at the bottom is the same density as before it was cooled, I thought I could use the value I calculated of the pressure before cooling and then determine the height.

    I had 1,081,990 Pa before cooling at the bottom so if I subtract the ice cap's pressure, I get 1,073,000.2 Pa.

    Subtract atmospheric pressure of 101,325 Pa to get 971,675.2 Pa. Then 971,675.2 = ρgh = 1000 * 9.80665 * h which gives a height of 99.08m. Resulting depth would be 100.08m (fresh water).

    I'm quite unsure about this. The result makes sense, as water expands when it freezes, but it doesn't feel like the proper way of solving the problem. I'm not sure how else to tackle it though.

    Any help is appreciated. Thanks
  2. jcsd
  3. Mar 30, 2014 #2


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    Staff: Mentor

    I am not sure about (e), but I don't see why pressure in (f) would change at all. The only thing that counts is mass of the column (approximated by ρh for a uniform column), and that didn't change.
  4. Mar 30, 2014 #3
    The way the problems progress suggest that you would use your answer for (e) to calculate the pressure at the bottom for (f). Something like 101325 + (1*916.7*9.80665) + ( whatever the answer from (e) for height was * 1000 * 9.80665) which would result in the pressure. That's why I'm unsure about my answer for (e). I have a lecture on it today so I will ask my lecturer and post his response.
  5. Mar 31, 2014 #4


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    Staff: Mentor

    Well, for sure you can use sum of ρgh for water and ρgh for ice to calculate the pressure, it is just completely unnecessary. ρgh is nothing else but mass times acceleration, and neither changes, regardless of how the density is distributed.

    For a very high column g would be a function of distance from the center of the Earth, but that's not the case here.
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