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Difference between in situ and potential temp in water

  1. Oct 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Given the compressibility of water, how do I find the temp change of a water parcel brought to the surface adiabatically? I know that it is equivalent to work done by the parcel, but am stuck as how to move forward. Given a depth, is there an equation to work out the temp change?


    2. Relevant equations
    compressibility=-1/V(dV/dp) and volumetric thermal expansion coefficient=1/V(dV/dT)


    3. The attempt at a solution
    I tried to use the first equation to get vol change and then vol change for temp change, but my answers are far too high. I assumed a unit volume. For a depth of 5000m the difference should be around half a degree. I used a conversion of around 1 bar to 10m.
     
  2. jcsd
  3. Oct 24, 2009 #2

    Mapes

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    Hi EMakepeace, welcome to PF. What term are you evaluating? What state variable remains constant when a process occurs adiabatically and reversibly?
     
  4. Oct 24, 2009 #3
    I am trying to derive an equation for the difference between potential temperature and in situ for a parcel of water. I realise that during the adiabatic process, there is no energy exchange between the parcel and the environment so that the change in thermal energy of the parcel equals just the work done as there is no heat exchange. I have only been given a value of compressibility (4.7x10^-5 bar^-1) and need to find the difference in temperature as the parcel expands with decreasing pressure/depth, as it is raised to the surface.

    I'm sorry for rambling but this is my first post and am unsure to how all this works!
     
  5. Oct 24, 2009 #4

    Mapes

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    There are a couple different ways to tackle the problem. If you're given the compressibility value, it's probably a good idea to work from there. How would you express the energy change [itex]dE[/itex] in terms of temperature change, and also in terms of work done? Does the fact that you know the relationship between [itex]dV[/itex] and [itex]dP[/itex] help you?

    This is a problem where attention to units will make or break you. Track your units carefully.
     
  6. Oct 24, 2009 #5
    I think I would relate dE to dT by dE=McdT, but this is where I am unsure. I can work out the change in volume from compressibility=-1/V(dV/dp), but then how to relate it to dE?
     
  7. Oct 24, 2009 #6

    Mapes

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    What is [itex]dE[/itex] in terms of work?
     
  8. Oct 24, 2009 #7
    dE=W(work done by the parcel)+Q(=0, in this case); so dE=W=PdV?
     
  9. Oct 24, 2009 #8

    Mapes

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    Great; try integrating this, knowing what you know about [itex]dV[/itex] from the compressibility relation.
     
  10. Oct 24, 2009 #9
    Many, many thanks!!
     
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