Tungsten filament-finding length of coil

1. Oct 29, 2007

laural

1. The problem statement, all variables and given/known data
Tungsten wire has an emissivity of 0.35. A tungsten filament in a light bulb can be considered to emit light only from the sides of the filament (consider it a cylinder... they are usually wire "cylinders" wrapped into coils.). If the operating temperature of the filament of a small 18W light bulb is 2800 K and the diameter of the filament is 0.25 mm, then what is the length of the coil (in cm)? Assume the inside of the light bulb is under vaccuum, and that the ends of the filament do not contribute any heat radiation, since they are attached to the circuit...and assume other energy loss in the circuit is negligible.

2. Relevant equations

I will be using & to represent delta as I am unsure how change in is represented on keyboard.

&Q/&t = ekAT^4 where k is the Stefan-Boltzmann constant and = 5.67 x 10^-8 W/m^2K^4

3. The attempt at a solution

I plugged in 18W as the rate of change of heat per time, and the provided emissivity for e

Plugged 2800 K in for change in temperature.

So:

18W= (0.35)*(5.67 x 10^-8 W/m^2K^4
)*A*(2800^4 K)

Solved A= 1.48x10^-5

Then used A=(pi)r^2h,

And solved h= 0.30 cm

I have been struggling with this problem and am not sure if this is right. I would appreciate it if someone would review my work.

Thank-you.

Edit: I realized I was using the wrong formula. Changed.

Last edited: Oct 29, 2007
2. Oct 30, 2007

rl.bhat

Check the calculation.
h = 1.48x10^-5/pix(0.125x10^-3)^2