Temperature of filament using its area, power, and emissivity

1. Feb 26, 2013

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.

2. Feb 26, 2013

CWatters

26 or 0.26 ?

3. Feb 26, 2013

Fluxthroughme

$e = 0.26 = \frac{26}{100}$. 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.

4. Feb 27, 2013

CWatters

Sorry that was the only possible error I could see.

5. Feb 27, 2013

Fluxthroughme

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 :)

6. Feb 27, 2013

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.

7. Feb 28, 2013

Fluxthroughme

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?

8. Feb 28, 2013

CWatters

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

9. Feb 28, 2013

haruspex

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.