How Does Water Density Change at Depths?

AI Thread Summary
The discussion focuses on calculating the change in water density at a depth of 400 meters in a lake, starting with a surface density of 1030 kg/m3 and using the bulk modulus of water. The formula used is rho = rho0(1/(1-dP/B)), where dP is the change in pressure. Participants discuss the challenge of calculating dP due to the non-constant density of water and suggest approximating it by assuming constant density initially. The final calculated density is approximately 2.12618, which is close to the expected value in the book. The method used, while not entirely accurate, is deemed acceptable for practical purposes given the pressure values involved.
Abhishekdas
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Change in density of water...

Homework Statement


Calculate the approximate change in density of water in a lake at a depth of 400m below the surface. The density of water at the surface is 1030kg/m3 and bulk modulus(B) of water is 2*109...

Homework Equations


rho=rho0(1/(1-dP/B))
rho = density of water at any depth
rho0 = density of water at surface
dP= change in pressure

The Attempt at a Solution


How to calculate dP if i use h*rho*g , rho is not constant...So how do i go about it...Please help...
 
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I only know how to approximate this without calculus...it gets very,very close to the real value anyway.
First, assume that the density remains constant, and find the pressure at 400m using that density. Then recalculate the second density using that pressure.
Hope this helps!
 


Ok...thanks i will try it...
I am sorry i didnt notice your reply for a long time...
 


my answer is 2.12618 and answer in the book is 2...I guess they are expecting this answer ...But you it should be close as we are dealing with values of pressure musch lesser than Bulk modulus of water...But this is'nt a correct method is'nt it? Neway...thanks a lot for your help...
 

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