Question Regarding Bulk Modulus

[Ithryndil]

The deepest point in any ocean is in the Mariana Trench, which is about 11 km deep, in the Pacific. The pressure at this depth is huge, about 1.13 X 10^8 N/m^2.

(a) Calculate the change in volume of 1 m3 of seawater carried from the surface to this deepest point.
wrong check mark m3

(b) The density of seawater at the surface is 1.03 X 10^3 kg/m3. Find its density at the bottom.
kg/m3

(c) Is it a good approximation to think of water as incompressible?
Yes
No

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The above is the question. Now, I didn't think I would have an issue with this, but apparently I am. The bulk modulus equation is:

B = -ΔP/ [(ΔV/Vi)] where:
B = bulk modulus, ΔP = change in pressure, ΔV = change in volume and Vi = initial volume.

I am trying to find ΔV for part A (afterwards the rest should not be difficult at all).

So ΔV = -[ΔP(Vi)]/B

When you use 0.21 X 10^10 for the bulk modulus of water (which apparently the online assignment website wants this for water and not seawater), you get:

ΔV = -(1.13 X 10^8)(1)/(0.21 X 10^10) = -0.0538 m^2

I actually inputted this into the website as a positive value and it counted it as wrong. The reason this is strange is because it will normally say "wrong sign" and if it's off by 10% or less it will say so. So I am hesitant to input in the negative value because that will be my last submission. Any help is appreciated. Thanks.

 

[Ithryndil]

This can be locked, I figured out the problem.

 

[Moetasim]

It's possible that the website is looking for the change in volume in cubic meters, not square meters. In that case, you would need to multiply your answer by the initial volume, 1 m^3. So the change in volume would be -0.0538 m^3.

Regarding the density at the bottom, you can use the equation p = p0/(1+βΔP), where p is the density at the bottom, p0 is the density at the surface, β is the compressibility of water (5.1 X 10^-10 Pa^-1), and ΔP is the change in pressure. Plugging in the values, you would get a density of 1.04 X 10^3 kg/m^3.

As for whether it is a good approximation to think of water as incompressible, it depends on the context and level of accuracy needed. Water is considered incompressible in most everyday situations, but at extreme depths and pressures, it does exhibit some compressibility. So it may be a good approximation for this particular question, but in general, it is not completely accurate.