Energy and Joule heating

by BillJ3986
Tags: energy, heating, joule
 P: 7 1. The problem statement, all variables and given/known data A solar panel (an assemblage of solar cells) measures 58 cm x 53 cm. When facing the sun, this panel generates 2.7A at 14V. Sunlight delivers an energy of 1.0 x 10^3 W/m^2 to an area facing it. What is the efficiency of this panel, that is, what fraction of the energy in sunlight is converted into electric energy? 2. Relevant equations Am I doing this problem correctly 3. The attempt at a solution The power delivered by the panel is P(delivered)= I x V= 2.7A x 14V= 37.8 W For the power delivered by sunlight I multiplied 1.0 x 10^3 W/m^2 by the area of .58m x .53m: so the area equals .307m^2. 1.0 x 10^3 W/m^2 x .307m^2= 307.4 W delivered by the sun. So I needed to find the fraction of energy in sunlight that is converted into electric energy. I did that by dividing 37.8W/307.4. The fraction of energy converted is .123 or 12.3%. Or is the answer 307.4/37.8= 8.13? I'm not sure which is the right answer or if it I did it correctly please let me know.
Mentor
P: 40,260
 Quote by BillJ3986 So I needed to find the fraction of energy in sunlight that is converted into electric energy. I did that by dividing 37.8W/307.4. The fraction of energy converted is .123 or 12.3%.
Looks good to me.

 Or is the answer 307.4/37.8= 8.13?
No, you want the fraction of the energy in the sunlight that is converted to electrical energy. Your first answer was correct. (Efficiency could never be greater that 1 or 100%,)
 P: 7 Thank you.