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The average American uses electrical energy at the rate of about 1.5 kW. Solar energy reaches Earth’s surface at an average rate of about 300 watts on every square meter (a value that accounts for nights and clouds). What fraction of the United State’s land area would have to be covered with 20% efficient solar cells to provide all of our electricity?

Respuesta :

gcosarinsky

Answer: the fraction is [tex]0.88*10^{-6}[/tex]

Explanation:

Hi!

The land area per capita in USA is 28,310 square km. Then the solar power per capita is (rounding some numbers):

[tex]P_s = 300 \frac{W}{m^2} (28,310) km^2 =  3*10^2*2.83*10^4*10^6 W = 8.49*10^{12} W = 8.49*10^9 kW[/tex]

If we take 20% of this power, the fraction k to have 1.5 kW is:

[tex]1.5 kW = k(0.2*8.49 * 10^9 kW)[/tex]

[tex]k = \frac{1.5}{1.7}*10^{-6} = 0.88 *10^{-9} [/tex]

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