(a) Calculate the power per square meter reaching Earth's upper atmosphere from the Sun. (Take the power output of the Sun to be 4.00 × 10 26 W.) (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m 2 reaches Earth's surface. Calculate the area in km 2 of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States' energy needs ( 1.05 × 10 20 J ) ? Australia's energy needs ( 5.4 × 10 18 J ) ? China's energy needs ( 6.3 × 10 19 J ) ? (These energy consumption values are from 2006.)
(a) Calculate the power per square meter reaching Earth's upper atmosphere from the Sun. (Take the power output of the Sun to be 4.00 × 10 26 W.) (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m 2 reaches Earth's surface. Calculate the area in km 2 of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States' energy needs ( 1.05 × 10 20 J ) ? Australia's energy needs ( 5.4 × 10 18 J ) ? China's energy needs ( 6.3 × 10 19 J ) ? (These energy consumption values are from 2006.)
(a) Calculate the power per square meter reaching Earth's upper atmosphere from the Sun. (Take the power output of the Sun to be
4.00
×
10
26
W.) (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m2 reaches Earth's surface. Calculate the area in km 2 of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States' energy needs
(
1.05
×
10
20
J
)
? Australia's energy needs
(
5.4
×
10
18
J
)
? China's energy needs
(
6.3
×
10
19
J
)
? (These energy consumption values are from 2006.)
The total rate at which power is used by humans worldwideis approximately 15 TW (terawatts). The solar flux averagedover the sunlit half of Earth is 680 W>m2 (assumingno clouds). The area of Earth’s disc as seen from the Sun is1.28 * 1014 m2. The surface area of Earth is approximately197,000,000 square miles. How much of Earth’s surfacewould we need to cover with solar energy collectors to powerthe planet for use by all humans? Assume that the solar energycollectors can convert only 10% of the available sunlightinto useful power.
What area is needed for a solar collector to absorb 45.0 kW ofpower from the Sun’s radiation if the collector is 75.0% efficient?(At the surface of Earth, sunlight has an average intensity of1.00 * 103W>m2.)
To heat a room with dimensions width a=3 m, length b=5 m, height h=2,2 m, approximately an electrical power of P=10 W per square meter is needed. At a cost of 0.2 soles per kW.h, how much will it cost per day to use this heater?
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