A home uses about 4*10^11 power joules of energy each year. In many parts of the United States there are about 3000 h of sunlight each year. How much energy from the sun falls on one square meter each year? If the solar energy can be converted to useful energy with an efficiency of 20%, how large an area of converters would produce the energy needed by the home?

Question

Jingles

Physics

26 January, 00:47

A home uses about 4*10^11 power joules of energy each year. In many parts of the United States there are about 3000 h of sunlight each year. How much energy from the sun falls on one square meter each year? If the solar energy can be converted to useful energy with an efficiency of 20%, how large an area of converters would produce the energy needed by the home?

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Francesca Hebert · Answer 1

The Sun transfers in each second and amount of 1367J per square meter, so in hour it can transfer:

E = 60*60*1367 J/h*m2 = 4,921,200 J/h*m^2.

in one year, we have 3000h*4,921,200 J/h*m^2. = 14,763,600,000 Joules per square meter.

Now, only 20% of this can be used as energy, so the amount that can be used is:

E = 0.2 * 14,763,600,000 J/m^2 = 2,952,720,000 J/m^2

Now, we want to find the number X of square meters such:

X*2,952,720,000 J/m^2 = 4x10^11 J

X = (4x10^11 J) / (2,952,720,000 J/m^2) = 135.5 m^2

So you only need around 135.5 square metters per house.