How Does a Tungsten Filament's Size Affect Its Radiated Power?

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The Tungsten filament of an incandescent lightbulb is a wire of diameter=0.080mm and length=5.0cm at a temperature of 3200K. What is the power radiated by the filament?


We Know:
Power=Watt/time=Energy/time'

E=h(nu)
nu=5.088E10 Hz/K

I'm not sure how to use the diameter and length to find the power. Since it's a filament, it would be in the shape of a cylinderso we could use the information to find the area, but I don't know how that applies.

Any help is appreciated!
 
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Quelsita said:
We Know:
Power=Watt/time

No. Power is measured in watts.

Quelsita said:
Power=Energy/time

Yes.


Quelsita said:
E=h(nu)
Yes, but not that relevant to the problem.


Quelsita said:
nu=5.088E10 Hz/K

How did you arrive at this number? No energy is given. Also, frequency is measured in just hertz. NOT hertz/kelvin.


As for the approach, have you heard of the Stefan-Boltzmann law?
 
cepheid said:
How did you arrive at this number? No energy is given. Also, frequency is measured in just hertz. NOT hertz/kelvin.



As for the approach, have you heard of the Stefan-Boltzmann law?

Sorry, it should be multplied by T which would cancel the Kelvin units.
I have but I was unsure what to do with the diameter and length information as they are not needed in the Stefan-Boltzmann law.

Since S-B law is power/unit area, could you just multiply the Sv by the area to get power?
 
Quelsita said:
Sorry, it should be multplied by T which would cancel the Kelvin units.

That's sort of beside the point. The unit error was not the main thing I was getting at. The main thing I was getting at was that E = hν is totally irrelevant to the problem! The fact that you have *somehow* mysteriously arrived at a number for ν makes absolutely no sense to me because 1. you don't have the information required in order to be able to calculate it (no energy is given) and 2. you are not *asked* to calculate it. I'm sorry, but your posts just aren't making a whole lot of sense.

Quelsita said:
I have but I was unsure what to do with the diameter and length information as they are not needed in the Stefan-Boltzmann law.
Ummm...really? You don't know what to do with that information? Continue reading...

Quelsita said:
Since S-B law is power/unit area, could you just multiply the Sv by the area to get power?

YES! Now tell me something...just how exactly do you expect to compute the area without the "diameter and length information." Now do you see why that is given?