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Chapter 1: Problem 3

Eight kilometres below the surface of the ocean the pressure is 81.7 MPa. Determine the density of sea-water at this depth if the density at the surface is \(1025 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) and the average bulk modulus of elasticity is \(2.34 \mathrm{GPa}\).

Short Answer

Expert verified
The density of sea-water at 8 km depth is approximately 1060.74 kg/m^3.

Step by step solution

01

Understand the problem

We need to determine the density of sea-water at a depth of 8 kilometers below the surface of the ocean using given values for surface density, pressure at depth, and the bulk modulus of elasticity.
02

Write down the given values

The given values are: surface density \( \rho_0 = 1025 \ \mathrm{kg} / \mathrm{m}^3 \), pressure at depth \( P = 81.7 \ \mathrm{MPa} \) and bulk modulus of elasticity \( K = 2.34 \ \mathrm{GPa} \).
03

Convert units

Convert the pressure and bulk modulus from MPa and GPa to Pa:\[ P = 81.7 \times 10^6 \ \mathrm{Pa} \]\[ K = 2.34 \times 10^9 \ \mathrm{Pa} \]
04

Use the bulk modulus formula

The bulk modulus formula is given by \[ K = -V \left(\frac{\Delta P}{\Delta V}\right) \]. Since density \( \rho \) is the inverse of volume, we can write \[ K = -\rho \left(\frac{\Delta P}{\Delta \rho}\right) \].
05

Solve for final density

To find the change in density, rearrange the equation:\[ K = \rho_0 \left(\frac{P}{\rho - \rho_0}\right) \]Solve for \( \rho \):\[ \rho = \rho_0 \left(1 + \frac{P}{K}\right) \]
06

Substitute the values into the equation

Substitute the given values into the density formula:\[ \rho = 1025 \left(1 + \frac{81.7 \times 10^6}{2.34 \times 10^9}\right) \]
07

Calculate the density

Perform the calculation:\[ \rho = 1025 \left(1 + 0.03487\right) = 1025 \times 1.03487 = 1060.74 \mathrm{kg/m}^3 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

bulk_modulus_of_elasticity
The bulk modulus of elasticity, denoted as K, is a measure of a substance's resistance to uniform compression. Essentially, it tells us how much the volume of a material decreases under pressure. The formula to express it is: \[ K = -V \left( \frac{\Delta P}{\Delta V} \right) \]. In our case, with seawater, we use an adapted version related to density: \[ K = -\rho \left( \frac{\Delta P}{\Delta \rho} \right) \]. The variables used are:
  • \( \rho \): Density of seawater
  • \( \Delta P \): Change in pressure
  • \( \Delta \rho \): Change in density
  • \( V \): Volume of seawater
This relationship is important in understanding how the seawater's density changes as we go to greater depths, where pressure is increased significantly.
pressure-depth_relationship
Pressure in a fluid increases with depth due to the weight of the fluid above. This is quantified by the pressure-depth relationship. In our problem, at 8 kilometers under the ocean's surface, the pressure is given as 81.7 MPa. To use this in calculations, we convert it to Pascals: \[ 81.7 \times 10^6 \ \text{Pa} \]. This relationship can be generalized as:
  • \( P = \rho_{fluid} \cdot g \cdot h \), where:
  • \(\rho_{fluid}\) is the density of the fluid at the surface
  • \(g\) is the acceleration due to gravity (approximately 9.81 m/s²)
  • \(h\) is the depth in the fluid
Our specific depth relationship helps us understand how the enormous pressures at significant ocean depths affect the physical properties of seawater, particularly its density.
density_calculation
The calculation of seawater density at depth relies on understanding how pressure affects density change through the bulk modulus. Using the formula derived for density: \[ \rho = \rho_0 \left(1 + \frac{P}{K}\right) \], we can substitute the given values:
  • \(\rho_0 = 1025 \ \text{kg/m}^3 \): Density of seawater at the surface
  • \(P = 81.7 \times 10^6 \ \text{Pa} \): Pressure at 8 km depth
  • \(K = 2.34 \times 10^9 \ \text{Pa} \): Average bulk modulus of elasticity for seawater
By substituting these into the equation, we get:
\[ \rho = 1025 \left(1 + \frac{81.7 \times 10^6}{2.34 \times 10^9}\right) \]
Performing the calculations gives the final density at depth:
\[ \rho \approx 1025 \times 1.03487 = 1060.74 \text{kg/m}^3 \]
Thus, the higher the pressure at greater depths compacts the seawater, increasing its density from 1025 kg/m³ at the surface to approximately 1060.74 kg/m³ at 8 km depth.

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Most popular questions from this chapter

At an absolute pressure of \(101.3 \mathrm{kPa}\) and temperature of \(20{ }^{\circ} \mathrm{C}\) the dynamic viscosity of a certain diatomic gas is \(2 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}\) and its kinematic viscosity is \(15 \mathrm{~mm}^{2} \cdot \mathrm{s}^{-1}\). Taking the universal gas constant as \(8310 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\) and assuming the gas to be perfect, calculate its approximate relative molecular mass.

A liquid of kinematic viscosity \(370 \mathrm{~mm}^{2} \cdot \mathrm{s}^{-1}\) flows through an \(80 \mathrm{~mm}\) diameter pipe at \(0.01 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\). What type of flow is to be expected?

By how much does the pressure in a cylindrical jet of water \(4 \mathrm{~mm}\) in diameter exceed the pressure of the surrounding atmosphere if the surface tension of water is \(0.073 \mathrm{~N} \cdot \mathrm{m}^{-1}\) ?

Calculate the density of air when the absolute pressure and the temperature are respectively \(140 \mathrm{kPa}\) and \(50{ }^{\circ} \mathrm{C}\) and \(\mathrm{R}=\) \(287 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)

A uniform film of oil \(0.13 \mathrm{~mm}\) thick separates two discs, each of \(200 \mathrm{~mm}\) diameter, mounted co-axially. Ignoring edge effects, calculate the torque necessary to rotate one disc relative to the other at a speed of \(44 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (7 rev/s) if the oil has a dynamic viscosity of \(0.14 \mathrm{~Pa} \cdot \mathrm{s}\).
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