Temperature of filament using its area, power, and emissivity

[Fluxthroughme]

I have entered the emissivity in the calculation so that we can treat it as a blackbody, allowing us to use I_{tot} = \sigma T^4. My book tells me the correct answer is 2.06*10^4, which I'd normally put down to a misprint, but if I use my value, I get a value for the next part which is a factor of 10 off. So either I'm wrong, or they followed through with an error of theirs?

Thanks for any help.

 

[CWatters]

26 or 0.26 ?

 

[Fluxthroughme]

26 or 0.26 ?

e = 0.26 = \frac{26}{100}. Which is why there is a 26 on the bottom and a second 100 on the top.

Edit: If you don't understand why I have put e there, or why it's even in the calculation, I'd be fine if you could help me achieve the answer via a different method.

 

[CWatters]

Sorry that was the only possible error I could see.

 

[Fluxthroughme]

Sorry that was the only possible error I could see.

Is this to say I should assume that the book has this answer wrong and, as a consequence, the other answer is wrong, too?

Either way, thank you :)

 

[haruspex]

Fwiw, a typical temperature for an incandescent light bulb would be around 3000K. At 20000K the tungsten would evaporate.
Btw, you didn't need to add a pi r squared term. The ends of the wire are not exposed.

 

[Fluxthroughme]

Fwiw, a typical temperature for an incandescent light bulb would be around 3000K. At 20000K the tungsten would evaporate.
Btw, you didn't need to add a pi r squared term. The ends of the wire are not exposed.

Sorry, I don't understand why I don't need the pi r squared term? Isn't the fillament just a cylinder, so the radiations comes from the whole of the surface area?

 

[CWatters]

Won't make much difference either way as the area of the ends is small.

 

[haruspex]

Sorry, I don't understand why I don't need the pi r squared term? Isn't the fillament just a cylinder, so the radiations comes from the whole of the surface area?

The ends of the filament will be the electrical connection into the circuit. They will leak a small amount of heat by conduction, perhaps more than would be lost by the same area through radiation, but you've no information on that.