Rough Estimate about electric light bulb

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Homework Help Overview

The discussion revolves around estimating the temperature of the filament in an electric light bulb emitting white light, along with estimating the surface area of the filament given a radiated power of 100 W.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the color of light and temperature, referencing Wien's Law and the Stefan-Boltzmann law. Questions arise about the validity of using a blackbody approximation and the implications of high temperatures on the light bulb's materials.

Discussion Status

Participants are actively engaging with the concepts, with some suggesting the use of Planck's law for a more accurate estimation. There is acknowledgment of the challenges in approximating the filament as a blackbody, and the discussion includes considerations of thermal dynamics within the bulb.

Contextual Notes

There is mention of the light bulb being in a partial vacuum, which influences heat dissipation mechanisms and the material properties of the glass in relation to the filament's temperature.

darkar
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Well the question asked us to make a rough estimate of the temperature of the filament in an ordinary electric light bulb which emits white light. And if it has radiated power of 100 W, make a rough estimate of the surface area of filament.

Well, the thing is how u make a rough estimation on the temperature?
 
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The color of the light determines the temperature of the radiating body, as described by Wien's Law. A body which is radiating white light has a peak wavelength of emission about in the middle of the visible spectrum.

After you know the temperature, you can use the Stephan-Boltzmann law to find the radiating area, knowing that the radiated power is 100 W.

- Warren
 
The gas pressure inside a typical light bulb is reduced so you can probably ignore thermal conduction as a dissipation mechanism and take the thermal energy as being radiated away - more or less as a blackbody.
 
Right, so assuming it is a black body, but will it be better to use Planck Spectral Distribution diagram to estimate it? But then the temperature is quite high, about 5500 K, is this possible? Really this hot?
 
That's about right, darkar. A blackbody at about 5500K produces white light. The Sun, for example, is such a blackbody. Approximating the light bulb filament as a blackbody is obviously an oversimplification, but it's all you can do in this context.

- Warren
 
Wow, didnt know the light bulb is that hot. But if that's so, wouldn't the glass melt?
 
darkar,

No, because the inside of a light bulb is a (partial) vacuum. As Tide says, the only mechanism by which heat can leave the filament is EM radiation, and glass is nearly transparent to the majority of the EM radiation thus produced.

- Warren
 
I see, thanks a lot everyone !
 
darkar said:
Wow, didnt know the light bulb is that hot.

And that is precisely why Thomas Edison had so much difficulty finding the right material and configuration for his lightbulb filaments! :)
 

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