Estimating Temperature and Wavelength of a Light Bulb Filament

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Homework Help Overview

The discussion revolves around estimating the operating temperature and peak wavelength of thermal radiation emitted by a tungsten filament in a 40W light bulb. Participants are exploring the implications of their assumptions and calculations related to the filament's dimensions and properties.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the Stefan-Boltzmann law for temperature estimation and the challenges of calculating wavelength based on energy considerations. There is also a focus on the importance of using the correct area in calculations and the implications of the filament's radius.

Discussion Status

Some participants have provided alternative temperature estimates and engaged in checking each other's calculations. There is an ongoing exploration of the filament's dimensions, with differing opinions on the radius affecting the temperature results. No consensus has been reached on the correct values, but there is productive dialogue regarding assumptions and calculations.

Contextual Notes

Participants are working with potentially ambiguous information regarding the filament's radius, which may be a source of error in their calculations. The original poster notes uncertainty about the specific heat capacity of tungsten and the implications of using a single temperature value for energy calculations.

Brewer
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I have a question which is:

The filament of a 40W light bulb has a radius of 1.5μm and a length of 10cm.
Estimate:
i) the operating temperature of the filament
ii) the peak wavelength of the emitted thermal radiation

Stating any assumptions made.

I assumed that the filament was made of tungsten and has an emissivity of 0.26.

I then used the formula:
H = AeσT^4

but this gave me a temperature of 139972.5K, which as far as I'm aware is hotter than the surface temperature of the sun right?

For the second part of the question, I wanted to use E=mcΔθ to get the energy, and then use this with E=hf to find the frequency and thus be able to calculate the wavelength. However after further consideration I decided against this idea, as I don't have a change in temperature, only a single value (and the lack of specific heat capacity of tungsten in the question also lead me to think that this wasn't the best course of action to take).

Any help would be greatly appreciated.
 
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I got about 7300 K for the temperature. Check your units and your math.
 
Thanks for that. After a more indepth read of my textbook I found I had to use the surface area of the filament, whereas I had been using the cross sectional area of it.

I think the value that I got in the end was 7325.9K

And for the wavelength I got about 409.5nm. Do you agree with that?
 
That's close - I get about 396 nm.
 
Brewer said:
I have a question which is:
The filament of a 40W light bulb has a radius of 1.5μm and a length of 10cm.
Estimate:
i) the operating temperature of the filament
ii) the peak wavelength of the emitted thermal radiation
Stating any assumptions made.
I assumed that the filament was made of tungsten and has an emissivity of 0.26.
I then used the formula:
H = AeσT^4
but this gave me a temperature of 139972.5K, which as far as I'm aware is hotter than the surface temperature of the sun right?
For the second part of the question, I wanted to use E=mcΔθ to get the energy, and then use this with E=hf to find the frequency and thus be able to calculate the wavelength. However after further consideration I decided against this idea, as I don't have a change in temperature, only a single value (and the lack of specific heat capacity of tungsten in the question also lead me to think that this wasn't the best course of action to take).
Any help would be greatly appreciated.
I think the radius of the wire is wrong. It cannot be [itex]\mu m[/itex]. That is way too thin. It might be 1.5 millimeters but that seems a little thick for a light filament. If you use 1.5 mm, the answer is about 1600 deg. K which is in the right order of magnitude.

AM
 
I agree it seems far too thin, but I'm working off the question I have infront of me. Its either 1.5 [itex]\mu m[/itex] or 15. There was a mark on the page just where you'd put a decimal point so whether its a decimal point or a photocopying error I don't know
 

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