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The total rate at which power used by humans worldwide is approximately \(15 \mathrm{TW}\) (terawatts). The solar flux averaged over the sunlit half of Earth is \(680 \mathrm{~W} / \mathrm{m}^{2}\). (assuming no clouds). The area of Earth's disc as seen from the sun is \(1.28 \times 10^{14} \mathrm{~m}^{2}\). The surface area of Earth is approximately \(197,000,000\) square miles. How much of Earth's surface would we need to cover with solar energy collectors to power the planet for use by all humans? Assume that the solar energy collectors can convert only \(10 \%\) of the available sunlight into useful power.

Short Answer

Expert verified
The area of Earth's surface that needs to be covered with solar energy collectors to power the planet for use by all humans can be found through the following steps: 1. Calculate the total solar power received by Earth's disc: \[ P_{total} = 680 \mathrm{~W/m^2} * 1.28 \times 10^{14} \mathrm{~m^2} \] 2. Calculate the power available after considering the efficiency of solar collectors: \[ P_{available} = P_{total} * 0.1 \] 3. Calculate the area of solar collectors required to generate the total power used by humans worldwide: \[ Area_{required} = \frac{15 \times 10^{12} \mathrm{~W}}{P_{available}} \] 4. Convert the area to square miles: \[ Area_{required~in~square~miles} = \frac{Area_{required}}{2.59 \times 10^6 \mathrm{~m^2/square~mile}} \] Calculating these values, we can determine the necessary area of Earth's surface to power the planet with solar energy collectors.

Step by step solution

01

Find the total solar power received by Earth's disc

First, let's find the total solar power received by Earth's disc from the sun by multiplying the solar flux by the area of Earth's disc. Remember that the solar flux is averaged over the sunlit half of Earth. \[ P_{total} = Solar~Flux * Area_{Earth's~disc} \] \[ P_{total} = 680 \mathrm{~W/m^2} * 1.28 \times 10^{14} \mathrm{~m^2} \]
02

Calculate the power available after considering the efficiency of solar collectors

Since we know that solar energy collectors can convert only 10% of the available sunlight into useful power, we need to multiply the total solar power received by the Earth's disc by this efficiency. \[ P_{available} = P_{total} * Efficiency \] \[ P_{available} = P_{total} * 0.1 \]
03

Calculate the area of solar collectors required to generate the total power used by humans worldwide

Now, we will calculate the area of solar collectors needed to generate enough power to meet the global human energy demand, which is 15 TW (or 15 × 10^12 W). To do this, we'll divide the total power used by humans (in watts) by the power available per square meter. \[ Area_{required} = \frac{Total~Power~used}{Power~per~m^2} \] \[ Area_{required} = \frac{15 \times 10^{12} \mathrm{~W}}{P_{available}} \]
04

Convert the area to square miles

Finally, to find the area in square miles, convert the calculated required area from square meters to square miles. Since 1 square mile is equal to \(2.59 \times 10^6\) square meters, you can divide the required area in square meters by this conversion factor. \[ Area_{required~in~square~miles} = \frac{Area_{required}}{2.59 \times 10^6 \mathrm{~m^2/square~mile}} \] Following these four steps will give us the area of Earth's surface that needs to be covered with solar energy collectors to power the planet for use by all humans.

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Key Concepts

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

Solar Power Calculation
Understanding how to calculate the power we can harness from the sun is pivotal in planning for a sustainable future. Solar power calculation is not just about acknowledging the abundance of sunlight, but also about figuring out the tangible means to convert this natural bounty into usable energy.

Let's dive into this with an example: imagine you have a solar panel right in front of you under the bright sun. The amount of power this panel can receive is determined by two factors: the intensity of the sunlight, known as 'solar flux', and the surface area of the panel exposed to the light. Solar flux, generally measured in watts per square meter (W/m²), varies depending on the time of day, weather conditions, and geographical location.

Now, taking the exercise we've been given—it provides us with a global solar flux of 680 W/m² on a cloudless day, averaged over Earth's sunlit half. By multiplying this flux by the area of Earth's disc as seen from the sun, we calculate the total solar power Earth receives on its daytime side. However, it is crucial to remember that not all this power can be turned into electricity—the efficiency of solar collectors comes into play, which leads us to our next concept.
Energy Conversion Efficiency
When looking to harness the sun's energy, 'energy conversion efficiency' is a term that frequently pops up. It quantifies the portion of energy that can be converted into a usable form compared to the total energy that is input—in our case, sunlight. Solar panels, unfortunately, do not have a 100% efficiency rate; a substantial amount of the sunlight's energy is lost.

Currently, most commercial solar panels have an efficiency ranging from 15% to 20%. For our textbook exercise, the assumed efficiency is 10%. This number is not arbitrary; it accounts for technological limitations and helps us estimate the realistic potential for energy capture. With this 10% efficiency, the vast amount of sunlight energy striking Earth's surface is significantly reduced in terms of the power available for our use.

From the given exercise, by utilizing the 10% efficiency rate, we can calculate the actual power available from our solar collectors. This step is key to understanding the ratio of utilized solar power to the power potential. Even with technological advancements, efficiency will always be a limiting factor but understanding it helps us optimize our energy systems for maximum benefit.
Renewable Energy Resources
In the bigger picture, solar energy is just one part of the tapestry of 'renewable energy resources' that include wind, hydro, geothermal, and biomass. The significance of these resources lies in their ability to be naturally replenished on a human timescale, presenting an enduring solution to meet the world's energy demands. Unlike fossil fuels, which are finite and emit harmful greenhouse gases, renewables provide a cleaner alternative.

The push for solar energy, as reflected in our exercise, is not only due to its abundance but because it is among the most environment-friendly options. Harnessing solar power helps in reducing carbon footprints and fighting climate change. When we discuss covering parts of Earth with solar collectors to meet our energy needs, it's not a mere hypothetical—it's part of a crucial strategy to shift towards sustainable living.

What's exciting is that every square meter of land can potentially become a power generator, contributing to a global energy network that is fed by the sun. With the continuous improvement in solar technology and increased governmental support for green energy, the focus on renewable resources has never been more important than now.

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Most popular questions from this chapter

A thief plans to steal a gold sphere with a radius of \(28.9 \mathrm{~cm}\) from a museum. If the gold has a density of \(19.3 \mathrm{~g} / \mathrm{cm}^{3}\), what is the mass of the sphere in pounds? [The volume of a sphere is \(V=(4 / 3) \pi r^{3}\).] Is the thief likely to be able to walk off with the gold sphere unassisted?

(a) The diameter of Earth at the equator is \(7926.381 \mathrm{mi}\). Round this number to three significant figures and express it in standard exponential notation. (b) The circumference of Earth through the poles is \(40,008 \mathrm{~km}\). Round this number to four significant figures and express it in standard exponential notation.

(a) After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A \(25.0\)-mL portion of the liquid had a mass of \(21.95 \mathrm{~g}\). A chemistry handbook lists the density of benzene at \(15^{\circ} \mathrm{C}\) as \(0.8787 \mathrm{~g} / \mathrm{mL}\). Is the calculated density in agreement with the tabulated value? (b) An experiment requires \(15.0 \mathrm{~g}\) of cyclohexane, whose density at \(25^{\circ} \mathrm{C}\) is \(0.7781 \mathrm{~g} / \mathrm{mL}\). What volume of cyclohexane should be used? (c) A spherical ball of lead has a diameter of \(5.0 \mathrm{~cm}\). What is the mass of the sphere if lead has a density of \(11.34 \mathrm{~g} / \mathrm{cm}^{3}\) ? (The volume of a sphere is \((4 / 3) \pi r^{3}\), where \(r\) is the radius.)

The density of air at ordinary atmospheric pressure and \(25^{\circ} \mathrm{C}\) is \(1.19 \mathrm{~g} / \mathrm{L}\). What is the mass, in kilograms, of the air in a room that measures \(14.5 \mathrm{ft} \times 16.5 \mathrm{ft} \times 8.0 \mathrm{ft}\) ?

Give the derived SI units for each of the following quantities in base SI units: (a) acceleration \(=\) distance \(/\) time \(^{2}\) (b) force \(=\) mass \(\times\) acceleration (c) work \(=\) force \(\times\) distance (d) pressure \(=\) force/area (e) power = work/time (f) velocity \(=\) distance/time (g) energy \(=\) mass \(\times(\text { velocity })^{2}\)

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