# Homework Help: Power of the sun

1. Sep 2, 2008

### shanie

1. The problem statement, all variables and given/known data
I'm carrying out a lab to investigate the power of the sun. I'm supposed to compare the light intensity of a 100W light bulb with the light intensity from the sun, by comparing the radiating heat on my cheeks (one cheek pointing to the sun, the other to the bulb). I found that the bulb had the same intensity as the sun from a distance of 0.1 m.

2. Relevant equations
A =4πr^2
IA=P

3. The attempt at a solution
area of
A=4πr2=0.126 m^2 from the bulb
with the effect 100W, gives the intensity,

I=P/A=100/0.126=795.8 lx

Which means that this is the intensity that radiates from the sun. Calculating the area for the sun, using the average measurement of 1.496*10^11 m from the Earth to the Sun:

A=4πr2=2.827*1023 m2
IA=P=795.8lx *2.827*1023 m2≈2.25*1026W=225 YW (yottawatt, SI)
Compared to the literary value 391 YW.

Is this calculation correct? And also, I explained the significant error between the theoretical value and the practical one by mentioning the fact that a lot of the sun's power is lost as it radiates in all directions and meets obstacles on its way to the Earth. In addition to the fact that the bulb probably doesn't give 100W in practice, but a lot less. Are these conclusions correct? I could really use some assistance, thanks!

2. Sep 2, 2008

### mgb_phys

First lux is not W/m^2 it's lumens/m^2 , but that's not important - the method is correct.
The usual value for the sun's irradiance at earth is around 1200w/m^2 at noon so 800 isn't too far out.

3. Sep 2, 2008

### LowlyPion

You might want to explain the error difference - i.e. how much measurement difference in your cheek from the bulb the resulting difference between published and calculated might be. (Simply reverse calculate using the published to determine what value would have yielded the correct answer. Is that distance then reasonable from your method?)

4. Sep 3, 2008

### shanie

Thank you! :)