How much visible light is emitted by a 100 W tungsten light bulb?

[fluidistic]

Homework Statement

Assume that the tungsten filament of an incandescent $$100 W$$ light bulb can be considered as a black body.
Estimate the percentage of the irradiated energy in the visible spectra.
The effective area of the filament is $$100 mm^2$$.
There are 3 more questions in the exercise that I'll try to do on my own.

Homework Equations

Not even sure.

The Attempt at a Solution

What I really think I need is the expression of the function $$u(\lambda )$$ that we can see in the picture http://en.wikipedia.org/wiki/File:Wiens_law.svg.
Otherwise than this, I've attempted to use Wien's displacement law to see what wavelength of light is most created by the filament. But it's not a good idea, I couldn't even make a percentage guess. For instance if I get a wavelength of the kilometer order (I know I should get infrared which is somehow shorter), I'd have no idea about what would be the percentage of energy irradiated in the visible spectra. I'm just out of ideas. I'd like to listen to you guys, what would you do?

 

[hikaru1221]

The spectral energy density: $$u(\lambda,T)=\frac{8\pi hc}{\lambda^5}\frac{1}{exp(\frac{hc}{\lambda k_BT})-1}$$
A typical incandescent light bulb has operating temperature of about 3000K. So one unknown is known
The visible spectrum is very narrow, i.e. from 400nm to 750nm. If you plot the function u on this interval, you will see that it's nearly a straight line. So the power of visible radiation emitted is:
$$P_{visible}=\int^{\lambda_{red}}_{\lambda_{blue}}Au(\lambda,T)d\lambda \approx A\frac{u(\lambda_{red})+u(\lambda_{blue})}{2}(\lambda_{red}-\lambda_{blue})$$

 

[fluidistic]

The spectral energy density: $$u(\lambda,T)=\frac{8\pi hc}{\lambda^5}\frac{1}{exp(\frac{hc}{\lambda k_BT})-1}$$
A typical incandescent light bulb has operating temperature of about 3000K. So one unknown is known
The visible spectrum is very narrow, i.e. from 400nm to 750nm. If you plot the function u on this interval, you will see that it's nearly a straight line. So the power of visible radiation emitted is:
$$P_{visible}=\int^{\lambda_{red}}_{\lambda_{blue}}Au(\lambda,T)d\lambda \approx A\frac{u(\lambda_{red})+u(\lambda_{blue})}{2}(\lambda_{red}-\lambda_{blue})$$

Thank you very much. Your function u reminded me of Plank's law, so I found in wikipedia your expression: http://en.wikipedia.org/wiki/Planck's_law#Overview.
I'll try to finish the exercise tomorrow, right now I'm going to bed.
Out of curiosity, you gave me an extra information, namely that the filament is at 3000K. But, shouldn't I derive it say by using the fact that it's a 100 W light bulb and that the effective area of the filament is 100 mm^2?

 

[Dick]

Thank you very much. Your function u reminded me of Plank's law, so I found in wikipedia your expression: http://en.wikipedia.org/wiki/Planck's_law#Overview.
I'll try to finish the exercise tomorrow, right now I'm going to bed.
Out of curiosity, you gave me an extra information, namely that the filament is at 3000K. But, shouldn't I derive it say by using the fact that it's a 100 W light bulb and that the effective area of the filament is 100 mm^2?

I would look at the Stefan–Boltzmann law. You've got 100W coming out of the area of the filament. Doesn't that let you estimate its temperature?

 

[hikaru1221]

Out of curiosity, you gave me an extra information, namely that the filament is at 3000K. But, shouldn't I derive it say by using the fact that it's a 100 W light bulb and that the effective area of the filament is 100 mm^2?

Thanks. Now you remind me of Stephan-Boltzmann law The first thing coming to my mind was an actual light bulb, so I used the 3000K temperature to estimate. Using Stephan-Boltzmann law would be more theoretically accurate, so I encourage you to apply the law
By the way, I didn't expect the theoretical temperature to be just 2000K (calculate and you will see it). Quite a big difference from practice.

P.S: That big difference is true. From this page: emissivity of tungsten aged filament is from 0.032 to 0.35, quite low but possible for metal.
http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html