What is the efficiency of a solar collector made of an aluminium canister?

[johnboy11]

Hi there,

As part of my A2 physics coursework investigation, i want to calculate the efficiency of an aluminium canister modeled as a solar collector.

The basic set up of the apparatus is:

- Lamp (40W light bulb) at a fixed distance of 5cm away from the aluminium canister (which is painted black, and the half of the canister not being penetrated by light is covered with insulating material - carpetty)

What measurements need to be taken? Can you provide me with a step-by-step guide of how to calculate the efficiency with equations.

Thanks a lot!

 

[neu]

You want a step by step guide of how to do your coursework with equations?

How about reading your recommended text and talking to your teacher?

 

[johnboy11]

No its not actually my coursework investigation, and isn't really required, but i want to do this anyway...

Its not in the textbooks, nor the A2 syllabus, but it shows that i have researched, which i have...

 

[neu]

No-one's going to give you a step by step guide of how to do an experiment. Ask a specific question about something you're unable to grasp.

Efficiency = useful power output / total power input

What is your useful output? Heat?

 

[mikelepore]

A common calculation related to the intensity of a light source is to figure that a detector gathers some of the source's energy on a particular surface area, and that area is a known fraction of the area of an imaginary sphere surrounding the source and capturing all of its energy. A common astronomy exercise is: a detector orbiting the Earth measures electromagnetic radiation from the sun to be 1370 watts per square meter, now, using a reference table to obtain the distance from the Earth to the sun, imagining the sun to be enclosed by a sphere whose radius is the sun-earth distance, we can determine that the sun puts a total of 3.8 X 10^26 watts.