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Chapter 8: Problem 129

The surface of the continental United States has an area of about \(8 \times 10^{6} \mathrm{~km}^{2}\) and an average elevation of about \(500 \mathrm{~m}\) (above sea level). The average yearly rainfall is \(75 \mathrm{~cm} .\) The fraction of this rainwater that returns to the atmosphere by evaporation is \(\frac{2}{3} ;\) the rest eventually flows into the ocean. If the decrease in gravitational potential energy of the water-Earth system associated with that flow could be fully converted to electrical energy, what would be the average power? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(1000 \mathrm{~kg} .)\)

Short Answer

Expert verified
The average power is approximately \(3.11 \times 10^{11}\) watts.

Step by step solution

01

Calculate Total Volume of Rainfall

First, we need to determine the total volume of rain that falls on the continental United States annually. The volume can be calculated as \[ \text{Volume} = \text{Area} \times \text{Rainfall Height} \]Substitute the given numbers:\[ \text{Volume} = (8 \times 10^6 \text{ km}^2) \times (0.75 \text{ m}) \]Convert the area from square kilometers to square meters:\[ 1 \text{ km}^2 = 10^6 \text{ m}^2\]So,\[ 8 \times 10^6 \text{ km}^2 = 8 \times 10^6 \times 10^6 \text{ m}^2 = 8 \times 10^{12} \text{ m}^2 \]Thus,\[ \text{Volume} = 8 \times 10^{12} \times 0.75 = 6 \times 10^{12} \text{ m}^3 \]
02

Calculate the Mass of Water that Flows into the Ocean

Given that \( \frac{1}{3} \) of the rainwater flows into the ocean, the mass can be calculated as:\[ \text{Mass} = \frac{1}{3} \times \text{Volume} \times 1000 \text{ kg/m}^3 \]Substitute the calculated volume:\[ \text{Mass} = \frac{1}{3} \times 6 \times 10^{12} \times 1000 = 2 \times 10^{15} \text{ kg} \]
03

Determine the Gravitational Potential Energy Decrease

We calculate the decrease in gravitational potential energy using the formula:\[ \Delta U = mgh \]where \( m = 2 \times 10^{15} \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( h = 500 \text{ m} \). Substituting these values gives:\[ \Delta U = 2 \times 10^{15} \times 9.81 \times 500 = 9.81 \times 10^{18} \text{ J} \]
04

Convert Energy to Average Power

To find the average power, we need to divide the total energy by the time period, which is one year. One year has approximately \( 365.25 \times 24 \times 3600 = 31,557,600 \text{ seconds} \).So the average power is:\[ P = \frac{9.81 \times 10^{18}}{31,557,600} \approx 3.11 \times 10^{11} \text{ W} \]

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Key Concepts

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

Rainfall calculation
To understand how much rainfall a large area like the continental United States receives, it's essential to calculate the total volume of rainfall over that area. The formula is quite straightforward:
  • Volume of rainfall = Area of the land x Rainfall height
Given that the U.S. land area is about \(8 \times 10^6\) km虏, and the average yearly rainfall is 75 cm or 0.75 meters, we first convert square kilometers into square meters:
  • \(1 \text{ km}^2 = 10^6 \text{ m}^2\)
Thus, \(8 \times 10^6 \text{ km}^2 = 8 \times 10^{12} \text{ m}^2\). By multiplying this by the rainfall height of 0.75 m:
  • Total volume = \(8 \times 10^{12} \times 0.75 = 6 \times 10^{12} \text{ m}^3\)
This calculation helps us gauge the vastness of the water involved in such a natural phenomenon.
Energy conversion
Energy conversion is the process of changing one form of energy into another, and it's crucial in studying how natural processes can be harnessed to produce useful work, such as electricity. In this scenario, we look at how gravitational potential energy is converted.
When water from rainfall makes its way to the ocean, it loses potential energy. This energy loss can theoretically be captured and converted to electrical energy.
  • Gravitational potential energy is calculated using \( \Delta U = mgh \)
  • Where \( m \) is the mass, \( g \) is the acceleration due to gravity (9.81 m/s虏), and \( h \) is the height (500 m).
If this energy could be fully converted into electricity, it could power a significant amount of electrical devices, illustrating the potential for clean energy generation.
Average power
Average power is all about the rate at which energy is converted from one form to another over time. It's calculated by dividing the total energy by the time period during which this energy transfer occurs.
  • Average power \( P = \frac{\text{Total Energy}}{\text{Time in seconds}} \)
  • In this case, we calculate the average power over one year, equivalent to about 31,557,600 seconds.
Using the decrease in gravitational potential energy \(9.81 \times 10^{18} \text{ J}\), the average power is found to be approximately \(3.11 \times 10^{11} \text{ W}\).
This high number demonstrates immense energy potential, comparable to several nuclear power plants operating in tandem.
Evaporation and water cycle
Evaporation plays a critical role in the water cycle, which is the natural process of water moving from the Earth's surface into the atmosphere and back again. The cycle is powered by the sun鈥檚 energy, and involves processes like
  • Evaporation: Water vapor rises from water bodies into the air.
  • Condensation: Vapor cools and forms clouds.
  • Precipitation: Water falls back to the Earth's surface as rain or snow.
  • Collection: Water gathers in bodies like rivers, lakes, and oceans.
In the U.S., \( \frac{2}{3} \) of the rainwater evaporates back into the atmosphere. This continuous cycle is crucial for replenishing water sources and maintaining ecosystems.
Mass and volume of water
Understanding the relationship between mass and volume of water is essential in many real-world applications. The density of water provides a reliable conversion factor because:
  • The density of water is approximately \(1000 \text{ kg/m}^3\).
This means for every cubic meter of water, it weighs around 1000 kg. For instance, if you calculate the volume of rain as \(6 \times 10^{12} \text{ m}^3\), the mass of this rainwater becomes \(6 \times 10^{12} \times 1000 = 6 \times 10^{15} \text{ kg}\).
By understanding this concept, one can appreciate the sheer weight of water involved in large-scale weather processes and their potential impacts on energy systems.

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Most popular questions from this chapter

A constant horizontal force moves a \(50 \mathrm{~kg}\) trunk \(6.0 \mathrm{~m}\) up a \(30^{\circ}\) incline at constant speed. The coefficient of kinetic friction is \(0.20 .\) What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

A \(20 \mathrm{~kg}\) block on a horizontal surface is attached to a horizontal spring of spring constant \(k=4.0 \mathrm{kN} / \mathrm{m} .\) The block is pulled to the right so that the spring is stretched \(10 \mathrm{~cm}\) beyond its relaxed length, and the block is then released from rest. The frictional force between the sliding block and the surface has a magnitude of \(80 \mathrm{~N}\). (a) What is the kinetic energy of the block when it has moved \(2.0 \mathrm{~cm}\) from its point of release? (b) What is the kinetic energy of the block when it first slides back through the point at which the spring is relaxed? (c) What is the maximum kinetic energy attained by the block as it slides from its point of release to the point at which the spring is relaxed?

A \(1.50 \mathrm{~kg}\) snowball is fired from a cliff \(12.5 \mathrm{~m}\) high. The snowball's initial velocity is \(14.0 \mathrm{~m} / \mathrm{s},\) directed \(41.0^{\circ}\) above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?

A collie drags its bed box across a floor by applying a horizontal force of \(8.0 \mathrm{~N}\). The kinetic frictional force acting on the box has magnitude \(5.0 \mathrm{~N}\). As the box is dragged through \(0.70 \mathrm{~m}\) along the way, what are (a) the work done by the collie's applied force and (b) the increase in thermal energy of the bed and floor?

In a circus act, a \(60 \mathrm{~kg}\) clown is shot from a cannon with an initial velocity of \(16 \mathrm{~m} / \mathrm{s}\) at some unknown angle above the horizontal. A short time later the clown lands in a net that is \(3.9 \mathrm{~m}\) vertically above the clown's initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?
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