# Homework Help: 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.