The solar power striking Earth every day averages 168 watts per square meter. The highest ever recorded electrical power usage in New York City was 13,200 MW. A record established in July of 2013. Considering that present technology for solar energy conversion is about 10% efficient, from how many square meters of land must sunlight be collected in order to provide this peak power? (For compar- ison, the total area of New York City is 830 km2.)

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Hey everyone in this example, we're told that the average annual electricity consumption of a household in florida is this value here per month. We need to assume that a solar panel has a 15.5% efficiency and a light from the sun applies 168 watts of energy per square meter. We need to calculate the area in meters squared of land that must be covered in order to achieve the average household need per month. We're going to assume that one month has 30 days. So our first step is to convert from kilowatt hours, two units of kilowatts. And we want to recall that for The amount of hours in a month. We would take 30 days in a month and multiply it by 24 hours, which is going to give us 720 hours. So taking our unit from the prompt, 1,141.50 kWh, we're going to convert to kW by dividing by 720 hours. And so this is going to give us our units in kW as 1.5854 kilowatts. So now converting from kilowatts, two units of watts. We're going to recall that our prefix kilowatt tells us that for one kilowatt we have 10 to the third power watts. We're now able to cancel out kilowatts and now we have the watts of energy required equal to 1585.4 watts. Now, according to the prompt, we have a solar panel with 15.5% efficiency and we're told that the light from the sun supplies 168 watts of energy per square meter. So we're going to use this information as a conversion factor, taking that percentage and writing it out as a decimal. We're going to take 1680.1 55 and we're going to multiply this by the watts of energy From the sun being 168 watts given in the prompt. And this is going to give us a value equal to 26.04 watts. And this is per meter squared according to the prompt per square meter. So so far here we have a conversion factor which is given from the prompt. We have here our watts of electricity required. And now we want to solve for the area that must be covered to achieve that energy need per household per month. So we would say that the area is equal to our watts required, which above we stated is 1585.4 watts. And because we recall from the prompt that our area should be in units of meters squared. We're going to convert from watts two m squared. So using this conversion factor that we've made above here, we can say that we have 26. watts for one m squared. And this is going to give us our final result where we will be able to cancel our units of watts. And sorry about that. So Watts is here, we're left with units of meters squared, and our final result is equal to 60.9 m squared. So this would be our final answer for our area that must be covered in order to achieve the average household energy need per month. So I hope that everything I viewed is clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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Key Concepts

Solar Power and Efficiency

Power Calculation

Area and Energy Density