Asked by Anonymous
Estimate the density of the water 6.0 deep in the sea. By what fraction does it differ from the density at the surface?
Answers
Answered by
drwls
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 http://en.wikipedia.org/wiki/Bulk_modulus
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.
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 http://en.wikipedia.org/wiki/Bulk_modulus
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.
Answered by
drwls
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.
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