Question

Solar radiation at the surface of the earth is about 700 W/m2. How much solar energy is incident on a roof of dimensions 8.83 m by 13.67 m in 4.43 hours?

Answer 1

Given:

The solar power per unit surface of the earth is: P = 700 W/m²

The dimensions of the roof are: A = 8.83 m × 13.67 m

The time for which the solar radiation is incident on the roof is: t = 4.43 h

To find:

The energy incident on the roof.

Explanation:

The area A of the roof is:

$A=8.83\text{ m}\times13.67\text{ m}=120.7061\text{ m}^2$

The time t can be converted into seconds as:

$t=4.43\text{ h}=4.43\times60\text{ min}=4.43\times60\times60\text{ s}=15.948\times10^3\text{ s}$

The energy incident on the given surface area in the given time is calculated as:

$E=P\times A\times t$

Substituting the values in the above equation, we get:

$\begin{gathered} E=700\text{ W/m}^2\times120.7061\text{ m}^2\times15.948\times10^3\text{ s} \\ \\ E=84.494\times10^3\times15.948\times10^3\text{ s} \\ \\ E=1347.51\times10^6\text{ W.s} \\ \\ E=1347.51\times10^6\text{ J} \\ \\ E=1347.51\text{ M.J} \end{gathered}$

Final answer:

The amount of solar energy incident on the roof is 1347.51 Mega Joules (M.J)