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The total consumption of electrical energy in the United States is about 1.0 \(\times 10^{19} \mathrm{J}\) per year. (a) What is the average rate of electrical energy consumption in watts? (b) The population of the United States is about 300 million people. What is the average rate of electrical energy consumption per person? (c) The sun transfers energy to the earth by radiation at a rate of approximately 1.0 \(\mathrm{kW}\) per square meter of surface. If this energy could be collected and converted to electrical energy with 40\(\%\) efficiency, how great an area (in square kilometers) would be required to collect the electrical energy used in the United States?

Short Answer

Expert verified
Average power: 3.17 × 10^11 watts, Per person: 1057 watts, Area: 792,500 km².

Step by step solution

01

Find the Average Power in Watts

To find the average power consumption in watts, we need to convert the given energy per year into joules per second (watts). The total consumption of electrical energy in the United States is 1.0 \times 10^{19} \text{ J/year}. Since power is energy per time and 1 year is 365 \times 24 \times 60 \times 60 seconds (or 31,536,000 seconds), we calculate:\[\text{Average power} = \frac{1.0 \times 10^{19} \text{ J}}{31,536,000 \text{ s}} \approx 3.17 \times 10^{11} \text{ watts}\]
02

Calculate Average Power Consumption Per Person

To find the average power consumption per person, we divide the total average power by the United States population:\[\text{Average power per person} = \frac{3.17 \times 10^{11} \text{ watts}}{300 \times 10^{6} \text{ people}} \approx 1057 \text{ watts/person}\]
03

Calculate Collected Energy Area with Efficiency

The solar energy incident on the surface of the Earth is 1.0 \text{kW/m}^2. With a 40\% efficiency, the usable power becomes \[1.0 \text{kW/m}^2 \times 0.4 = 0.4 \text{kW/m}^2\]We need to equate this to the average power consumption, so the required area \( A \) is given by:\[A = \frac{3.17 \times 10^{11} \text{ watts}}{0.4 \times 10^3 \text{ watts/m}^2} \approx 7.925 \times 10^8 \text{ m}^2\]Converting square meters to square kilometers:\[A \approx 792,500 \text{ km}^2\]

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Key Concepts

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

Average Power
Average power is a fundamental concept that helps us understand how much energy is used over a specific period. When we talk about electrical energy consumption, especially on a grand scale like that of an entire country, average power gives us insight into the consistent rate of energy use. The unit of power in electrical terms is the watt (W), which is equivalent to one joule per second.
To find the average power consumption, we take the total energy used over a year and divide it by the number of seconds in that year. In the context of the United States, which consumes about 1.0 \(\times 10^{19}\) joules annually, the average power is calculated by dividing this energy by the total seconds in a year (31,536,000 seconds). This calculation gives us approximately 3.17 \(\times 10^{11}\) watts.
  • Average Power is calculated as: \[\text{Average Power} = \frac{\text{Total Energy}}{\text{Time}}\]
  • Helps in understanding energy demands over time
  • Useful for planning and managing energy resources effectively
Population Energy Use
Understanding the average energy consumption per person is crucial when examining a population's energy footprint. It allows us to grasp how much energy, on average, each individual uses and can highlight variations across different regions or countries.
In the United States, dividing the total average power consumption by the population (around 300 million people) provides the average power used per individual. This amount comes to about 1057 watts per person. It can be a useful metric for comparing energy use with other countries or assessing the impact of various energy efficiency measures.
  • Average consumption per person illustrates personal energy footprint
  • Supports energy policy development and conservation strategies
  • Investigation into personal and societal habits impacting consumption
Solar Energy Efficiency
Solar energy is a promising alternative to traditional fossil fuels. Its potential is rooted in how much energy the sun delivers to the Earth's surface — roughly 1 kW per square meter. By harnessing this energy through solar panels, and assuming a conversion efficiency of 40%, it's possible to generate a significant amount of electrical power.
To replace the United States' consumption entirely with solar energy at 40% efficiency, a substantial area would be required to collect the necessary energy. Calculations show that to meet this demand, the collection area would need to be approximately 792,500 square kilometers. This highlights the practical challenges and scale needed for large-scale solar implementation.
  • Efficiency is key in converting solar radiation into usable energy
  • Requires significant area to meet large energy demands
  • Bridges the gap towards more sustainable energy production

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Most popular questions from this chapter

A Spring with Mass. We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M\) , equilibrium length \(L_{0}\) , and spring constant \(k\) . The work done to stretch or compress the spring by a distance \(L\) is \(\frac{1}{2} k X^{2}\) , where \(X=L-L_{0}\) (a) Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\) . Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of \(M\) and \(v\) . (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each piece in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L\) . The result is not \(\frac{1}{2} M v^{2},\) since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 \(\mathrm{kg}\) and force constant 3200 \(\mathrm{N} / \mathrm{m}\) is compressed 2.50 \(\mathrm{cm}\) from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 0.053 -kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

A Near-Earth Asteroid. On April \(13,2029\) (Friday the 13 th!), the asteroid 99942 Apophis will pass within \(18,600 \mathrm{mi}\) of the earth-about 1\(/ 13\) the distance to the moon! It has a density of 2600 \(\mathrm{kg} / \mathrm{m}^{3}\) , can be modeled as a sphere 320 \(\mathrm{m}\) in diameter, and will be traveling at 12.6 \(\mathrm{km} / \mathrm{s}\) . (a) If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver? (b) The largest nuclear bomb ever tested by the United States was the "CastlefBravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases \(4.184 \times 10^{15} \mathrm{J}\) of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?

Varying Coefficient of Friction. A box is sliding with a speed of 4.50 \(\mathrm{m} / \mathrm{s}\) on a horizontal surface when, at point \(P\) . it encounters a rough section. On the rough section, the coefficient of friction is not constant, but starts at 0.100 at \(P\) and increases linearly with distance past \(P\) , reaching a value of 0.600 at 12.5 \(\mathrm{m}\) past point \(P .(\text { a) Use the work-energy theorem to find how far this box }\) slides before stopping. \((b)\) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of 0.100\(?\)

A \(4.80-\mathrm{kg}\) watermelon is dropped from rest from the roof of a 25.0 \(\mathrm{m}\) -tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what is the watermelon's (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be different if there were appreciable air resistance?

Rescue. Your friend (mass 65.0 \(\mathrm{kg}\) ) is standing on the ice in the middle of a frozen pond. There is very hittle friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 \(\mathrm{s}\) and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) while you remain at rest. What is the average power supplied by the force you applied?

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