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.

Step-by-step explanation:

The area A of the roof is:

`A=8.83\text{ m}*13.67\text{ m}=120.7061\text{ m}^2`

The time t can be converted into seconds as:

`t=4.43\text{ h}=4.43*60\text{ min}=4.43*60*60\text{ s}=15.948*10^3\text{ s}`

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

`E=P* A* t`

Substituting the values in the above equation, we get:

`\begin{gathered} E=700\text{ W/m}^2*120.7061\text{ m}^2*15.948*10^3\text{ s} \\ \\ E=84.494*10^3*15.948*10^3\text{ s} \\ \\ E=1347.51*10^6\text{ W.s} \\ \\ E=1347.51*10^6\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)