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\(\bullet\) In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 \(\mathrm{km}\) and the pressure is \(1.16 \times 10^{8}\) Pa (about 1150 atmospheres). (a) If a cubic meter of water is taken to this depth from the surface (where the normal atmospheric pressure is about \(1.0 \times 10^{3} \mathrm{Pa}\) , what is the change in its volume? Assume that the bulk modulus for seawater is the same as for freshwater \(\left(2.2 \times 10^{9} \mathrm{Pa}\right) .\) (b) At the surface, seawater has a density of \(1.03 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) What is the density of sea-water at the depth of the Challenger Deep?

Short Answer

Expert verified
The change in volume cannot be calculated as the number of steps is incomplete. The first half is named and described promptly, but the second half is missing the necessary steps.

Step by step solution

01

Understand the Problem

We are given a depth at the Marianas Trench where the pressure is significantly higher than at the surface. We need to compute the change in volume of a cubic meter of seawater taken to this depth. We will use the bulk modulus of seawater, which is provided as being the same as freshwater.
02

Calculate the Change in Pressure

The change in pressure, \(\Delta P\), is the difference between the pressure at depth and the pressure at the surface. With the depth pressure as \(1.16 \times 10^{8} \, \text{Pa}\) and surface pressure as \(1.0 \times 10^{3} \, \text{Pa}\), we have: \(\Delta P = 1.16 \times 10^{8} - 1.0 \times 10^{3} = 1.15999 \times 10^{8} \, \text{Pa}\).

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Key Concepts

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

Bulk Modulus
The bulk modulus of a material like seawater is essential for understanding how it responds to pressure changes. It is a measure of material's resistance to compression. The bulk modulus can be thought of as a "stiffness" parameter for fluids.
In mathematical terms, the bulk modulus ( ) is given by the ratio of the change in pressure ( ) to the fractional change in volume ( ). This is expressed as:\[ K = -V \frac{\Delta P}{\Delta V} \]where is the initial volume, is the change in pressure, and is the change in volume.

The negative sign indicates that an increase in pressure leads to a decrease in volume. With seawater, we usually assume a bulk modulus similar to freshwater, which is about . Understanding this concept is crucial for predicting how seawater will behave under extreme pressures, like those found in the deep ocean.
Density of Seawater
Density is a fundamental property that tells us how much mass is packed into a given volume of substance. It's crucial when considering seawater at different depths. Seawater's density at the surface is , which is slightly higher than that of freshwater because of the salt content.
As we descend into deeper waters, the increased pressure impacts the water's density. This is because pressure compresses the water, slightly reducing its volume and thus increasing its density.

### Formula for Density ChangeTo calculate the new density ( ) at great depths like the Challenger Deep, we need to consider:
  • The original surface density ( )
  • The volumetric change under pressure ( )
The formula is typically:\[ \rho_\text{depth} = \frac{\rho_\text{surface} \times V_\text{surface}}{V_\text{depth}} \]Where and are the initial and final volumes. This enhancement in density is a direct consequence of the water compressing under pressure.
Volume Change
When you lower a cubic meter of seawater from the surface to extreme depths like those of the Marianas Trench, its volume changes due to the immense pressure increase. The bulk modulus helps calculate this change in volume.
### Calculating Volume ChangeUsing the relationship:\[ \Delta V = -\frac{V \Delta P}{K} \]where is the original volume of seawater and all other variables have been defined.
Inserting the given values from the exercise, we find out how much smaller the cubic meter of seawater becomes under the ocean's pressure.
Recognizing these changes can help us understand the physics of oceanic conditions and how they affect any object submerged to such depths.
Pressure at Depth
Pressure significantly increases with depth due to the weight of the water column above pushing down on the layers below. For each of depth, pressure increases by about . But at Challenger Deep, we're dealing with extreme depths, leading to much higher pressure values.
### Calculating Deep Sea PressureThe pressure at any depth ( ) is calculated considering the surface pressure and the height of the water column:\[ P = P_\text{surface} + \rho g h \]where is the surface pressure, is water density, is gravitational acceleration ( ), and is the depth.
This equation helps us predict the intense pressure of found in deep-sea environments like the Marianas Trench. By understanding this, we can design equipment to withstand these harsh conditions for scientific explorations.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) An apple weighs 1.00 \(\mathrm{N}\) . When you hang it from the end of a long spring of force constant 1.50 \(\mathrm{N} / \mathrm{m}\) and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back-and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed)?

\(\bullet\) A thin, light wire 75.0 \(\mathrm{cm}\) long having a circular cross section 0.550 \(\mathrm{mm}\) in diameter has a 25.0 \(\mathrm{kg}\) weight attached to it, causing it to stretch by 1.10 \(\mathrm{mm}\) . (a) What is the stress in this wire? (b) What is the strain of the wire? (c) Find Young's modulus for the material of the wire.

\(\bullet\) A proud deep-sea fisherman hangs a 65.0 \(\mathrm{kg}\) fish from an ideal spring having negligible mass. The fish stretches the spring 0.120 \(\mathrm{m} .\) (a) What is the force constant of the spring? (b) What is the period of oscillation of the fish if it is pulled down 3.50 \(\mathrm{cm}\) and released?

\(\bullet\) \(\bullet\) Human hair. According to one set of measurements, the tensile strength of hair is 196 \(\mathrm{MPa}\) , which produces a maximum strain of 0.40 in the hair. The thickness of hair varies considerably, but let's use a diameter of 50\(\mu \mathrm{m}\) . (a) What is the magnitude of the force giving this tensile stress? (b) If the length of a strand of the hair is 12 \(\mathrm{cm}\) at its breaking point, what was its unstressed length?

\(\bullet\) Effect of diving on blood. It is reasonable to assume that the bulk modulus of blood is about the same as that of water \((2.2 \mathrm{GPa}) .\) As one goes deeper and deeper in the ocean, the pressure increases by \(1.0 \times 10^{4}\) Pa for every meter below the surface. (a) If a diver goes down 33 \(\mathrm{m}\) (a bit over 100 \(\mathrm{ft} )\) in the ocean, by how much does each cubic centimeter of her blood change in volume? (b) How deep must a diver go so that each drop of blood compresses to half its volume at the surface? Is the ocean deep enough to have this effect on the diver?

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