Comparing Light Intensity: 100W Bulb to Sunlight

AI Thread Summary
The discussion focuses on comparing the light intensity of a 100W bulb with sunlight by measuring the heat on the cheeks from each source. The calculations show that the bulb's intensity at 0.1 m is approximately 795.8 lux, which is used to estimate the sun's total power output as around 225 yottawatts, compared to the known value of 391 yottawatts. The user acknowledges potential errors, noting that the bulb may not actually emit a full 100W and that sunlight's intensity diminishes as it travels through space. Additionally, the importance of understanding the difference between lux and watts per square meter is highlighted. Overall, the calculations and conclusions about the intensity comparison are deemed mostly correct, with suggestions for refining the error analysis.
shanie
Messages
23
Reaction score
0

Homework Statement


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.


Homework Equations


A =4πr^2
IA=P

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!
 
Physics news on Phys.org
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.
 
shanie said:
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.

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?)
 
Thank you! :)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top