Marianas Trench Homework: Volume & Density Calcs

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Homework Help Overview

The discussion revolves around a physics problem related to the Marianas Trench, specifically focusing on the effects of pressure on the volume and density of seawater at significant depths. The problem involves calculating the change in volume of seawater when taken from the surface to a depth of 10.9 km and determining the density of seawater at that depth.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between pressure, volume change, and density, with some exploring how to calculate density based on mass and volume. Questions arise regarding the relevance of initial conditions and how to apply the given equations.

Discussion Status

The discussion has progressed with participants sharing their reasoning and calculations. Some have provided guidance on how to approach the density calculation, while others are clarifying their understanding of the relationships between mass, volume, and density. There appears to be a productive exchange of ideas, though not all participants are in agreement on the approach.

Contextual Notes

Participants are working under the constraints of the problem statement, which specifies the initial conditions and parameters such as the density of seawater at the surface and the value of k for seawater. There is an acknowledgment of potential confusion regarding the initial volume being one cubic meter.

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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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