Question

The total amount of solar energy that arrives at the earth from the sun is 1.74 × 1017 W, and the surface area of the earth is 5.1 × 1014 m2 (assuming the earth's radius is 6.37 × 106 m).

(a) If a solar panel is about 20% efficient in converting solar photons into electrical power, how big a solar panel (in area, m2) would be required to run a 70 W light bulb?

(b) Humans use approximately 4.87 × 1018 J of electrical energy per year. What fraction of the available solar energy is required to meet this consumption?

Answer 1

solar energy/m2=1.74*10¹⁷W/(5.1*10¹⁴) =341 W/m2

considering 20% efficiency of solar energy, energy that can be transferred =341*0.2 = 68.2 W/m2

wattage of bulb= 70W

hence area required= 70/68.2 =1.03 m2

2. Energy available= 1.74*10¹⁷*0.2 = 3.48*10¹⁶ J/sec

year= 365 days =365*24hr*60min/hr*60sec/min=3153600 sec

energy available in a year= 3153600*3.48*10¹⁶ J= 1.097*10²⁴ Joules

energy consumed by humans= 4.87*10¹⁸ Joules in a year

hence fraction of energy required= 4.87*10¹⁸/ (1.097*10²⁴)= 4.43*10^-6