Question

The intensity of sunlight incident on the surface of a solar panel on your roof is 1000 W/m^2. Calculate the sunlight energy absorbed by the panel in one hour if the length of the panel is 6.00 m and its width is 4.00 m and the sunlight is incident perpendicularly on the panel.

Answer 1

8.64 x 10^7 J

Step-by-step explanation:

Intensity of sunlight, I = 1000 W/m^2

Length of panel, L = 6 m

Width of panel, W = 4 m

Area of the panel, A = L x W = 6 x 4 = 24 m^2

time, t = 1 hour = 3600 second

Energy = Intensity x area of panel x time

E = 1000 x 24 x 3600 = 8.64 x 10^7 J