Sunlight of intensity 1300 W/m squared falls on a collector of area 0.500 m squared for 1.00 h. How much energy does the collector receive?

1. Intensity is power per unit area, and power is energy per unit time, so I = (E / t) / A. 2. Rearranging for energy gives E = I A t. 3. Convert the time: 1.00 h = 3600 s. 4. E = (1300 W/m squared)(0.500 m squared)(3600 s). 5. First, (1300)(0.500) = 650 W, the power intercepted by the collector. 6. Then E = (650 W)(3600 s) = 2.34 x 10^6 J. 7. Stopping at 650 W reports a power as though it were an energy, which is the trap in one option. 8. Using 1.00 h as though it were 1 s leaves 6.50 x 10^2 J, three thousand times too small. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section 16.11 Energy in Waves: Intensity_