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Bulk modulus, water density

  1. May 22, 2005 #1
    An incerase of pressure of one atmosphere, causes a reduction of [tex]49*10^-^6m^3[/tex] in a volume of [tex]1m^3[/tex] Give the value of the bulk-modulus of water. Calculate the denisty of the water of a lake, at a depth of 200 metres, assuming that the density of water is [tex]1000kgm^-^3[/tex] at the surface.


    Calculating the bulk modulus was pretty easy, gives a value of [tex]2.04*10^9Nm^-^2[/tex], but i dont really know how to relate this to a change in the denisty over a given depth... seems like an integration over a rate of change with density, but the question cant involve doing that. confused.. :\ help appreciated.
  2. jcsd
  3. May 22, 2005 #2


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    Why can't the question involve an integration? The density is a function of the pressure. The pressure is a function of the weight of water above the depth, which involves an integral over depth of the density.

    Construct a little cylinder of height dh and area A with volume Adh containing a weight of water = g*density*volume. What is the pressure change between the surfaces of the cylinder?
  4. May 22, 2005 #3
    theres no way of finding an area...? besides, this class hasn't involved, or even been taught integration so i dont figure it would make sense to stick an integral into the class questions. :uhh:
  5. May 22, 2005 #4


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    The area is arbitrary. You could make it anything you want. It divides out of the calculation. If integration is out of the question, you can still do an approximate calculation. Assume the density is constant, and calculate the pressure at a depth of 200 meters. That can be done without integration. Then calculate the density of water at that depth. If you want to refine the approximation, you could use that result to find an average density between the surface and a depth of 200 meters to recalculate the pressure at 200 meters depth, then recalculate the density. I think you will find that the second step does not make very much difference, so the first level of approximation is close enough.

    To find the pressure at 200m, pick any area you want to use (e.g., 1 m^2) and calculate the weight of water in a cylinder (or rectangular prism, or prism with any shape cross section- the shape does not matter) with that area and 200m height. Then divide by the area to find the pressure.
    Last edited: May 22, 2005
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