# Density of water using bulk modulus

1. Sep 3, 2014

### toothpaste666

1. The problem statement, all variables and given/known data
Estimate the density of the water 5.3km deep in the sea. (bulk modulus for water is B=2.0×109N/m2.)

2. Relevant equations
P = F/A = Dgh

3. The attempt at a solution

I dont even know how to go about starting this one. Can someone point me in the right direction?

2. Sep 3, 2014

### Staff: Mentor

Yes. They want you to start out with the density of water at the surface, and, based on the pressure at depth and the bulk modulus, calculate the density at depth. Don't forget that the density is the reciprocal of the specific volume.

Chet

3. Sep 3, 2014

### toothpaste666

so D=1000kg/m^3 at surface. P at depth = Dgh = (M/V)gh and B for water = 2.0x10^9. My main problem is I cant see what to calculate the mass or volume of. And I cant think of how the bulk modulus would tie in because B = -P/(V/V0) and once again I know nothing about any volume. I am completely stuck on this one =[

4. Sep 3, 2014

### AlephZero

If you think about it physically, the density would be the same for any volume of water, so long as the volume is small compared with the whole of the sea.

So try using your equations starting with an arbitrary volume $V_0$. Do it using algebra. Don't start plugging in numbers till you have a "formula" for the answer.

5. Sep 3, 2014

### SteamKing

Staff Emeritus
1. The oceans are not composed of fresh water.
2. If you have a cube of seawater whose volume is 1 m^3 at the surface, what would the volume of this cube be at a depth of 5.3 km?

6. Sep 4, 2014

### Staff: Mentor

Or alternately, take as a basis 1 kg of mass. What would its volume be at the surface? If the pressure on this same kg of mass were raised to that at depth, what would be its new volume?

Chet

7. Sep 4, 2014

### toothpaste666

ok so this is what i have so far:
v' = change in velocity P'= change in pressure.
denisty of seawater = 1025 kg/m^3
assuming denisty D is constant
v'/v = P'/B = (Dg(h2)-Dg(h1))/B
h1 is the height of surface which is 0 so

v'/v = Dg(h2)/B = (1025)(9.8)(5300)/(2.0*10^9) = .027

so v'/v = .027
if the volume of the cube of water was 1m^3
v'=.027
so the change in volume was .027.
D= m'/v'
we are using the same mass and finding the difference in volume. at the surface D=1025 and V=1m^3 so M=1025kg
so D=m/v' = 1025/.027 = 38000

8. Sep 4, 2014

### toothpaste666

its says i am wrong. I am very stuck on this one =[

9. Sep 4, 2014

### SteamKing

Staff Emeritus
Isn't v' the change in volume, not velocity?

You are saying that the change in volume, not the new volume, is 0.027 m^3

You have to use the new volume of the seawater at depth, rather than the change in volume, to calculate the new density. You seem to have gotten your formulas mixed up.

10. Sep 4, 2014

### toothpaste666

the volume was compressed because the mass is the same and density is higher so
1-.027 = .973
the new volume is .973
D= 1025/.973 = 1053 kg/m^3

11. Sep 4, 2014

### toothpaste666

which is correct. thank you all :)

12. Sep 4, 2014

### Staff: Mentor

The left hand side of your equation should be (v'/v)-1, not v'/v.

Chet

13. Sep 5, 2014

### toothpaste666

out of curiosity, how would I find a precise answer to this? I would guess setting up an integral but since the density and depth are both variant does that mean I would need a double integral?

14. Sep 5, 2014

### Staff: Mentor

Not exactly. You would write down a pair of coupled ordinary differential equations involving the density ( or specific volume) and the pressure as dependent variables. These equations would have to be solved simultaneously.

Chet