Estimate the density of the water

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 11K views
kritzy
Messages
12
Reaction score
0

Homework Statement


Estimate the density of the water 5.7 km deep in the sea. (bulk modulus for water is B=2.0 x 10^9 N/m^2) By what fraction does it differ from the density at the surface?

Homework Equations


P=(rho)gh=F/A=B(delta l/l)

The Attempt at a Solution


So I have several equations above. I wanted so solve for rho using the first equation but I don't know pressure. I tried the second equations but I don't know Force or area. I tried the last equation but I would need delta l and l so basically I'm stuck. Some advice would be much appreciated.
 
Physics news on Phys.org
kritzy said:
Estimate the density of the water 5.7 km deep in the sea. (bulk modulus for water is B=2.0 x 10^9 N/m^2) By what fraction does it differ from the density at the surface?

Homework Equations


P=(rho)gh=F/A=B(delta l/l)

The Attempt at a Solution


So I have several equations above. I wanted so solve for rho using the first equation but I don't know pressure. I tried the second equations but I don't know Force or area. I tried the last equation but I would need delta l and l so basically I'm stuck. Some advice would be much appreciated.

Well they do say estimate. So maybe try to work it out assuming that p doesn't change?

(Yes it changes, but does it change enough to matter? See what results you get and then decide.)
 
LowlyPion said:
Well they do say estimate. So maybe try to work it out assuming that p doesn't change?

(Yes it changes, but does it change enough to matter? See what results you get and then decide.)

I don't understand. I'm suppose to be solving for p at a certain density. How can it not change?
 
kritzy said:
I don't understand. I'm suppose to be solving for p at a certain density. How can it not change?

Have you calculated it using a uniform p as to the effect it will have on a bulk modulus of 2 * 109?
 
In 6 km, the change in pressure is significant.

Here's something to get you started:

Use the equation,

B=dP/(d(rho)/rho)

Manipulate and integrate,

integral of (d(rho)/rho) = integral of (dP/B)

That results in,

ln(rho2/rho1)=exp((P2-P1)/B)

Where you can say state 1 is the surface, and state 2 is the state at 5.7 km down.

You can then use the rho*g*h equation.