Marianas Trench Homework: Volume & Density Calcs

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In the discussion on calculating volume and density in the Challenger Deep of the Marianas Trench, participants address two main problems. For part A, the change in volume of seawater at a depth of 10.9 km is calculated using the seawater compressibility factor, resulting in a decrease of -5.3 x 10^-2 m. In part B, the challenge lies in determining the density of seawater at this depth, given its surface density of 1.03x10^3 kg/m^3. The solution involves calculating the mass from the initial volume and then dividing it by the adjusted volume after accounting for the change. The clarity of the initial volume being 1 m³ is emphasized as crucial for solving the density problem.
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Homework Statement


In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 km and the pressure is 1.16x10^8 Pa (about 1.15x10^3 atm.

A) If a cubic meter of water is taken from the surface to this depth, what is the change in its volume? (Normal atmospheric pressure is about 1.0x10^5 Pa. Assume that k for seawater is 45.8x10^-11 Pa^-1

B)What is the density of seawater at this depth? (At the surface, seawater has a density of 1.03x10^3 kg/m^3


Homework Equations


DeltaP/K


The Attempt at a Solution


Well for part A I multiplied the K of seawater by the pressure, and got the right answer which is -5.3 x 10^-2 m.
Now I'm kind of stumped on part B. I feel like it's incredibly easy but don't really know where to begin. Obviously m/v=d, but I'm not sure if that's relevant here. Don't know how to relate pressure or depth with density.
 
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You are given density at the surface, and you calculated how much the volume would change if you moved a certain mass of water down. Keep in mind, that volume of water had a mass, and it did not change. Therefore you have the quantities needed to calculate density.
 
Alright well let's see.

mass divided by initial volume is 1030.
mass divided by initial volume + delta volume = the density I'm looking for.

Is this the right track?
 
My first instant was that since mass is negligent, the change in volume should equal the change in density. But that doesn't seem to be working.
 
javacola said:
Alright well let's see.

mass divided by initial volume is 1030.
mass divided by initial volume + delta volume = the density I'm looking for.

Is this the right track?

Yes. You also know that the initial volume is 1m^3, because that's what the problem says. You can easily find the mass, and from there, the final density.
 
Mass divided by initial volume is 1030. Use this, since you know initial volume, to find mass. Take this mass and divide it by initial volume - delta volume to find the density you are looking for.
 
Thanks a lot guys. Got it with 3 minutes to spare. Have no clue how you knew initial volume was 1, that's what threw me off.
 
javacola said:
Thanks a lot guys. Got it with 3 minutes to spare. Have no clue how you knew initial volume was 1, that's what threw me off.

We know that because the question states:

"If a cubic meter of water is taken..."
 
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