Density of water using bulk modulus

[toothpaste666]

Homework Statement

Estimate the density of the water 5.3km deep in the sea. (bulk modulus for water is B=2.0×109N/m2.)

Homework Equations

P = F/A = Dgh

The Attempt at a Solution

I don't even know how to go about starting this one. Can someone point me in the right direction?

 

[Chestermiller]

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

 

[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 can't see what to calculate the mass or volume of. And I can't 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 =[

 

[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.

 

[SteamKing]

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 can't see what to calculate the mass or volume of. And I can't 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 =[

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?

 

[Chestermiller]

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

 

[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

 

[toothpaste666]

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

 

[SteamKing]

ok so this is what i have so far:
v' = change in velocity P'= change in pressure.

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

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.

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

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

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.

 

[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

 

[toothpaste666]

which is correct. thank you all :)

 

[Chestermiller]

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

Chet

 

[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?

 

[Chestermiller]

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