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U.S. power use. The total consumption of electrical energy in the United States is about \(1.0 \times 10^{19}\) joules per year. (a) Express this rate in watts and kilowatts. (b) If the U.S. population is about 320 million people, what is the average rate of electrical energy consumption per person?

Short Answer

Expert verified
The U.S. uses about \(3.17 \times 10^{11}\) watts or \(3.17 \times 10^{8}\) kilowatts; each person uses ~991 watts on average.

Step by step solution

01

Convert Energy Use to Watts

Electrical energy consumption in the United States is given as \(1.0 \times 10^{19}\) joules per year. To express this in watts, we first convert the time period to seconds (since 1 watt = 1 joule/second). There are 365 days/year, 24 hours/day, 60 minutes/hour, and 60 seconds/minute, resulting in \(365 \times 24 \times 60 \times 60 = 31,536,000\) seconds in a year. Now calculate the power in watts: \[ \text{Power (P)} = \frac{1.0 \times 10^{19} \text{ joules}}{31,536,000 \text{ seconds}} \approx 3.17 \times 10^{11} \text{ watts} \]
02

Convert Watts to Kilowatts

To convert from watts to kilowatts, divide by 1,000 since 1 kilowatt = 1,000 watts. Thus: \[ 3.17 \times 10^{11} \text{ watts} = \frac{3.17 \times 10^{11}}{1000} \approx 3.17 \times 10^{8} \text{ kilowatts} \]
03

Calculate Energy Consumption per Person

The U.S. population is approximately 320 million people. To find the average rate of electrical energy consumption per person, divide the total power by the population. The calculation in watts per person is: \[ \frac{3.17 \times 10^{11} \text{ watts}}{320 \times 10^{6}} \approx 991 \text{ watts per person} \]

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Key Concepts

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

Power Conversion
Power conversion is the process of changing energy from one unit or form to another. In the context of electrical energy, we often convert energy units such as joules into power units like watts. This conversion is crucial to understand how much energy is used over time.
For example, knowing the total energy consumption in the U.S. is expressed in joules per year. However, to understand it as a power rate, we need to convert it into watts. Power is defined as energy per unit of time, so we need a common time unit, usually seconds, to convert joules into watts. Understanding power conversion helps in evaluating and comparing energy efficiencies and consumption patterns across different systems and periods.
  • Why Convert? Makes consumption understandable and comparable.
  • Unit Reference: 1 watt equals 1 joule per second.
  • Time Factor: Converts yearly data into seconds for watt conversion.
Joules to Watts
Converting joules to watts is essential when you need to describe energy consumption as a power rate. A joule is a measure of energy, whereas a watt is a measure of power, which is energy per unit time. The conversion uses the formula:
\[ \text{Power (W)} = \frac{\text{Energy (J)}}{\text{Time (s)}} \]To convert the U.S. annual energy consumption into watts, we first determine the total seconds in a year. Since the U.S. consumes approximately \(1.0 \times 10^{19}\) joules per annum, dividing this by the total seconds in a year gives the power consumption in watts. This conversion provides a clearer, more tangible understanding of energy usage over the year.
  • Unit Conversion: From joules (energy) to watts (power).
  • Time Translation: Establishes power by incorporating time.
  • Practical Use: Simplifies comparison of energy consumption.
Kilowatts Conversion
Kilowatts conversion is the next step after converting energy to watts. A kilowatt is 1,000 watts and is used to simplify large power ratings. It gives a more manageable figure compared to using just watts. For instance, U.S. energy consumption figures may appear enormous in watts but converting them to kilowatts can provide clarity.
Dividing the power in watts by 1,000 efficiently translates it into kilowatts. This is particularly useful in daily applications, such as in electric bills and appliance power ratings, as they are usually stated in kilowatts.
  • Reduction: Simplifies large watt figures.
  • Accessibility: Improves understanding for household references.
  • Common Practice: Used in billing and appliance specs.
Average Energy Consumption
Average energy consumption refers to the typical amount of energy used by an individual, household, or population over a certain period. Here, we evaluate the average energy usage by U.S. residents. By dividing the total energy power in watts by the population, we find how much an average person consumes.
This breakdown into average per capita consumption is crucial for understanding individual responsibility and footprint concerning national energy consumption. It shows how much energy, on average, each person uses, shedding light on both energy efficiency and potential areas for conservation.
  • Per Person Calculation: Important for personal energy impact.
  • Insightful Metrics: Helps target energy-saving initiatives.
  • Scale: Refines massive data into individual relatable insights.
U.S. Electrical Energy Use
U.S. electrical energy use encompasses the total amount of electricity consumed by all sectors in the nation. It is a crucial indicator of economic activities, and it affects local and global environmental policies. Understanding how much energy the U.S. uses helps gauge its energy demands and can influence sustainable practices.
Significant portions of U.S. energy consumption come from industrial, commercial, residential, and transportation sectors, reflecting the nation's development and technology use. Analyzing electrical energy use helps in devising strategies for energy conservation, efficiency improvements, and policy formulations aimed at reducing carbon footprints.
  • National Scale: Reflects the entire country's utilization.
  • Sector Breakdown: Shows contributions from different economic activities.
  • Environmental Impact: Guides towards eco-friendly policy making.

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Most popular questions from this chapter

All birds, independent of their size, must maintain a power output of \(10-25\) watts per kilogram of body mass in order to fly by flapping their wings. (a) The Andean giant hummingbird (Patagona gigas) has mass \(70 \mathrm{~g}\) and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) A \(70 \mathrm{~kg}\) athlete can maintain a power output of \(1.4 \mathrm{~kW}\) for no more than a few seconds; the steady power output of a typical athlete is only \(500 \mathrm{~W}\) or so. Is it possible for a human-powered aircraft to fly for extended periods by flapping its wings? Explain.

A fisherman reels in \(12.0 \mathrm{~m}\) of line while landing a fish, using a constant forward pull of \(25.0 \mathrm{~N}\). How much work does the tension in the line do on the fish?

Marbles of mass \(m\) are thrown from the edge of a vertical cliff of height \(h\) at speed \(v_{0} .\) Neglecting air resistance, how fast (in terms of \(m, h,\) and \(v_{0}\) ) will these marbles be moving when they reach the bottom of the cliff if they are thrown (a) straight up, (b) straight down, or (c) horizontally away from the cliff? Will the final velocity vectors of the marbles be the same or different for each case?

On flat ground, a \(70 \mathrm{~kg}\) person requires about \(300 \mathrm{~W}\) of metabolic power to walk at a steady pace of \(5.0 \mathrm{~km} / \mathrm{h}(1.4 \mathrm{~m} / \mathrm{s})\) Using the same metabolic power output, that person can bicycle over the same ground at \(15 \mathrm{~km} / \mathrm{h}\). A \(70 \mathrm{~kg}\) person walks at a steady pace of \(5.0 \mathrm{~km} / \mathrm{h}\) on a treadmill at a \(5.0 \%\) grade (that is, the vertical distance covered is \(5.0 \%\) of the horizontal distance covered). If we assume the metabolic power required is equal to that required for walking on a flat surface plus the rate of doing work for the vertical climb, how much power is required? A. \(300 \mathrm{~W}\) B. 315 W C. \(350 \mathrm{~W}\) D. \(370 \mathrm{~W}\)

A \(61 \mathrm{~kg}\) skier on level snow coasts \(184 \mathrm{~m}\) to a stop from a speed of \(12.0 \mathrm{~m} / \mathrm{s} .\) (a) Use the work-energy theorem to find the coefficient of kinetic friction between the skis and the snow. (b) Suppose a \(75 \mathrm{~kg}\) skier with twice the starting speed coasted the same distance before stopping. Find the coefficient of kinetic friction between that skier's skis and the snow.

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