1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Light bulbs as black bodies

  1. Sep 8, 2010 #1

    fluidistic

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Assume that the tungsten filament of an incandescent [tex]100 W[/tex] 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 [tex]100 mm^2[/tex].
    There are 3 more questions in the exercise that I'll try to do on my own.
    2. Relevant equations

    Not even sure.

    3. The attempt at a solution

    What I really think I need is the expression of the function [tex]u(\lambda )[/tex] 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?
     
  2. jcsd
  3. Sep 8, 2010 #2
    The spectral energy density: [tex]u(\lambda,T)=\frac{8\pi hc}{\lambda^5}\frac{1}{exp(\frac{hc}{\lambda k_BT})-1}[/tex]
    A typical incandescent light bulb has operating temperature of about 3000K. So one unknown is known :biggrin:
    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:
    [tex]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})[/tex]
     
  4. Sep 8, 2010 #3

    fluidistic

    User Avatar
    Gold Member

    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?
     
  5. Sep 8, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  6. Sep 9, 2010 #5
    Thanks. Now you remind me of Stephan-Boltzmann law :biggrin: 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 :smile:
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Light bulbs as black bodies
  1. Black body (Replies: 3)

  2. Light Bulb (Replies: 3)

  3. Light bulb (Replies: 4)

  4. Light bulb (Replies: 1)

  5. Light bulbs (Replies: 3)

Loading...