## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The area of the solar collector is $41.7~m^2$
Let $A$ be the area of the solar collector. The total energy collected by the solar collector is the intensity multiplied by the area multiplied by the time. Therefore; $I~A~t = E$ $A = \frac{E}{I~t}$ $A = \frac{150\times 10^6~J}{(1000~W/m^2)(3600~s)}$ $A = 41.7~m^2$ The area of the solar collector is $41.7~m^2$.