Post a New Question


posted by .

Estimate the density of the water 6.0 deep in the sea. By what fraction does it differ from the density at the surface?

  • Physics -

    What you need to answer this question is the bulk modulus of liquid water, K. It is the ratio of the pressure to the relative change in density that is caused by that pressure.

    K = P/(drho/rho)
    where "rho" is the density and "drho" is the change.

    The value of K, and a better explanation can be found at
    K is about 2.2*10^9 Pascals.

    At H = 6 meters depth in the ocean, the pressure is
    given by P = (rho)*g*H
    and rho = 1.00*10^3 kg/m^3

    The relative density change is therefore
    d(rho)/rho = 10^3*9.8*6.0/2.2*10^9
    or about 25 parts per million.

  • Physics (added comment) -

    The value of P that I used is actually the pressure increase above ambient pressure, and that is the value that should be used in this case.

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question