Question

The total annual U.S. energy consumption is about 2×1020 joules.

With current technologies and solar collectors on the ground, the best we can hope for is that solar cells will generate an average (day and night) power of about 180 watts/m2 . What total area would we need to cover with solar cells to supply all the power needed for the United States? Give your answer in both square meters and square kilometers.

Answer 1

We know, 1 watt = 1 J/s

Now, power output in 1 day or 24 hr = 180 W/m^2

So, energy output in 24 hr = (180*24*3600) J/m^2

= 15552000 J/m^2

Thus energy output in 1 year or 365 days will be

= (15552000*365) J/m^2 = 5676480000 J/m^2

That is, for 5676480000 J, we need = 1m^2

For 1 J, we need = 1/5676480000 m^2

For 2*10^20 J, we need = (2*10^20/5676480000)

= 3.523*10^10 m^2

Now, 1 km^2 = 10^6 m^2 => 1 m^2 = 10^-6 km^2

So, we need = (3.523*10^10*10^-6) km^2

= 3.523*10^4 km^2