Answer to Question #130210 in Physics for Elvis

Question

The total energy consumption of a country is of the order 108 J. What area of solar collector would be needed to produce this energy, given that the power falling on the Earth's surface due to solar radiation is 1400 W/m2. Assume 100% efficiency. Density of water = 1000kg/m3
Weight of 1 m3 of water= 9800 N
Average solar energy flux= 500 W/m2

Answer 1

Assume that in a year a country consumes 10⁸ J of energy. How many watts is this? Divide by the number of seconds in a year:

$$P=\frac{10^8}{365\cdot24\cdot3600}=3.17\text{ W}.$$

With an average solar density flux in the country of 500 W/m², the area needed to cover this energy need is

$$A=\frac{P}{p}=\frac{3.17}{500}=6.34\cdot10^{-3}\text{ m}^2.$$