Density of Sea Water at 1000m Depth

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Homework Help Overview

The discussion revolves around determining the density of seawater at a depth of 1000 meters, where the pressure is approximately 1.0 x 10^7 pascal. The original poster provides the surface density of seawater as 1.03 x 10^3 kilograms per cubic meter and references the bulk modulus of water.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use the bulk modulus to relate pressure changes to volume changes and subsequently to density changes. Some participants question the assumption that depth can be treated as a force, clarifying that depth is a distance and discussing hydrostatic pressure.

Discussion Status

Participants are actively engaging in clarifying concepts related to pressure, depth, and the bulk modulus. Some guidance has been offered regarding the calculation of pressure at depth and the use of the bulk modulus to estimate density changes, with suggestions for further refinement through integration.

Contextual Notes

There is an emphasis on using the bulk modulus in the solution, and participants are navigating the constraints of assumptions about constant density and pressure in their calculations.

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What the density of sea water at depth of 1000 M where the water pressure is about 1.0*10^7 pascal the density of sea water at the surface is 1.03*10^3 kilograms per cubic M?


Homework Equations


BULK MODULUS=dP/dV/V
p=pressure
v=volume
B of water=,21*10^10 n/m^2
it is approx to the bulk of sea water


The Attempt at a Solution


Ihave used the (1000m)depth as aforce at the bottom then A=f/p=1000/1.0*10^7=10^-4 M^2
dV=A*h=10^-4*1000=10^-1 m^3
from bulk modulus=dp/dv/v=,21*10^10=10^7/dv/v then dv/v=4,76*10^-3 then v(surface)= x then density surface=mass/v then mass = x then v depth = dv+v surface(because as account to the bulk modulus the volume smaller at the depth) then the density of the depth= mass/v depth

it right to assume that the depth is aforce?
 
Last edited:
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no, depth is not a force, depth is a distance

a supported column of fluid in a gravitational field leads to a hydrostatic pressure (which has units of Force/Area)

the hydrostatic pressure in a constant gravitational field, is given by
[tex]P = \rho g h[/tex]
where h is the depth
 
thank you for help but g is the acceleration due gravity ?what p
can you help me mor is there another method ?but I have to use the bulk modulus in the solution I will be very grateful for your help

thank you a lot
 
Last edited:
g is the acceleration due to gravity

so, first calculate the pressure at 1000m assuming the density is constant

then assume the bulk modulus is constant with pressure & calculate the corresponding volume & density change. (similar to what you attempted previously)

This will be a pretty good approximatino to the densety change, to do any better you would have to set up an integral
 

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