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In the Challenger Deep of the Marianas Trench, the depth of seawater is \(10.9 \mathrm{~km}\) and the pressure is \(1.16 \times 10^{8} \mathrm{~Pa}\) (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 \(\left.1.0 \times 10^{5} \mathrm{~Pa}\right),\) 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 seawater at the depth of the Challenger Deep?

Short Answer

Expert verified
a) Change in volume is -0.0527 m^3. b) Density at depth is 1087 kg/m^3.

Step by step solution

01

Calculate the Change in Pressure

To find the change in volume (\(\Delta V\),) of the water, we first need to find the change in pressure (\(\Delta P\),). The surface pressure is given as 1.0 \(\times\) 10^5 Pa, and the pressure at the depth is 1.16 \(\times\) 10^8 Pa. Therefore, the change in pressure is:\[\Delta P = P_{\text{depth}} - P_{\text{surface}} = 1.16 \times 10^8 - 1.0 \times 10^5 = 1.159 \times 10^8 \, \text{Pa}.\]
02

Use Bulk Modulus to Find Change in Volume

The bulk modulus (\(B\),), given as 2.2 \(\times\) 10^9 Pa, relates pressure change to volume change by the formula:\[B = -\frac{\Delta P}{\frac{\Delta V}{V_0}},\]where \(V_0\) is the initial volume, and \(\Delta V\) is the change in volume. Rearranging the formula gives:\[\Delta V = -\frac{\Delta P \times V_0}{B}.\]Substituting in the values (\(V_0\) = 1 m^3):\[\Delta V = -\frac{1.159 \times 10^8 \times 1}{2.2 \times 10^9} = -0.0527 \, \text{m}^3.\]
03

Calculate Final Volume at Depth

The final volume (\(V\),) at depth is given by:\[V = V_0 + \Delta V = 1 - 0.0527 = 0.9473 \, \text{m}^3.\]
04

Calculate Density at Depth

To find the density at the depth, use the formula:\[\rho = \frac{m}{V},\]where \(m\) is the mass and \(V\) is the volume. Since the mass remains constant, its surface density is given by \(\rho_0 = 1.03 \times 10^3 \, \text{kg/m}^3\). Thus, the volume at the surface is still 1 m^3 and the mass is:\[m = \rho_0 \times V_0 = 1.03 \times 10^3 \times 1 = 1.03 \times 10^3 \, \text{kg}.\]
05

Calculate Final Density

At depth, the volume changes, so the density becomes:\[\rho' = \frac{m}{V} = \frac{1.03 \times 10^3}{0.9473} \approx 1087 \, \text{kg/m}^3.\]

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Key Concepts

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

Bulk Modulus
The bulk modulus is a measure of a material's resistance to uniform compression. It describes how easy or difficult it is to change the volume of a substance when under pressure. It is defined by the formula: \( B = -\frac{\Delta P}{\frac{\Delta V}{V}} \), where \( \Delta P \) is the change in pressure, \( \Delta V \) is the change in volume, and \( V \) is the original volume.
It is important to note that the bulk modulus is a positive value, meaning the more pressure you apply, the less a material compresses. For seawater, the bulk modulus is typically about \( 2.2 \times 10^9 \) Pa. This value helps us understand how seawater behaves at different depths, especially when dealing with phenomena such as deep-sea diving or undersea habitats.
Understanding the bulk modulus is crucial in fluid mechanics because it assists in predicting how fluids will behave under various pressure conditions, including those in the deep ocean floor.
Challenger Deep
The Challenger Deep is the deepest known point in Earth's seabed hydrosphere. Located in the Mariana Trench in the western Pacific Ocean, its depth is about 10.9 km. This dramatic depth means extraordinary pressures, vastly different from anything found on land.
For context, the pressure at Challenger Deep is around \( 1.16 \times 10^8 \) Pa, 1150 times the atmospheric pressure at sea level. Such extreme conditions make it a significant study area for fluid mechanics, allowing scientists to explore and deepen their understanding of pressure effects on materials.
Exploring these deep parts of the ocean poses extreme challenges due to the substantial pressure and has led to the development of advanced materials and technologies that can withstand such environments.
Density of Seawater
The density of seawater is a crucial concept in oceanography and fluid mechanics. At the surface, seawater typically has a density of about \( 1.03 \times 10^3 \text{ kg/m}^3 \). However, this density increases with depth due to the increase in pressure, caused by the weight of the overlying water.
In the Challenger Deep, the pressure causes the seawater's volume to decrease, resulting in a higher density. This phenomenon is calculated using the mass of seawater and its volume at depth, with the changes in volume resulting from physical compression according to pressure principles in fluid mechanics.
Understanding the density variations in seawater is critical for navigation, submarine buoyancy control, and studying ocean currents and marine life habitats.
Change in Pressure
The change in pressure, \( \Delta P \), is a pivotal factor when analyzing fluid mechanics, especially in deep-sea environments. It represents the difference between the pressure experienced at the surface of the ocean and the pressure at a significant depth, such as the Challenger Deep.
To calculate \( \Delta P \), we subtract the surface pressure from the deep-sea pressure. For example, starting with a surface pressure of \( 1.0 \times 10^5 \) Pa and a depth pressure of \( 1.16 \times 10^8 \) Pa, the pressure difference is \( \Delta P = 1.159 \times 10^8 \) Pa.
This increase in pressure at greater depths impacts the behavior of fluids, causing compression that can significantly affect the volume and density of the submerged material. Analyzing these pressure changes allows us to predict how materials and structures behave under deep-sea conditions, from engineering applications to biological studies.

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Most popular questions from this chapter

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 \(1.0 \times 10^{4} \mathrm{~Pa}\) 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?

A steel wire \(2.00 \mathrm{~m}\) long with circular cross section must stretch no more than \(0.25 \mathrm{~cm}\) when a \(400.0 \mathrm{~N}\) weight is hung from one of its ends. What minimum diameter must this wire have?

Stress on a mountaineer's rope. A nylon rope used by mountaineers elongates \(1.10 \mathrm{~m}\) under the weight of a \(65.0 \mathrm{~kg}\) climber. If the rope is \(45.0 \mathrm{~m}\) in length and \(7.0 \mathrm{~mm}\) in diameter, what is Young's modulus for this nylon?

A concrete block is hung from an ideal spring that has a force constant of \(200 \mathrm{~N} / \mathrm{m} .\) The spring stretches \(0.120 \mathrm{~m} .\) (a) What is the mass of the block? (b) What is the period of oscillation of the block if it is pulled down \(1.0 \mathrm{~cm}\) and released? (c) What would be the period of oscillation if the block and spring were placed on the moon?

A steel wire has the following properties: $$ \begin{array}{l} \text { Length }=5.00 \mathrm{~m} \\ \text { Cross-sectional area }=0.040 \mathrm{~cm}^{2} \end{array} $$ Young's modulus \(=2.0 \times 10^{11} \mathrm{~Pa}\) Shear modulus \(=0.84 \times 10^{11} \mathrm{~Pa}\) Proportional limit \(=3.60 \times 10^{8} \mathrm{~Pa}\) Breaking stress \(=11.0 \times 10^{8} \mathrm{~Pa}\) The proportional limit is the maximum stress for which the wire still obeys Hooke's law (see point \(\mathrm{B}\) in Figure 11.12 ). The wire is fastened at its upper end and hangs vertically. (a) How great a weight can be hung from the wire without exceeding the proportional limit? (b) How much does the wire stretch under this load? (c) What is the maximum weight that can be supported?

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