The cylindrical filament in a light bulb has a diameter of 0.050 mm, an emissivity of 1.0, and a temperature of 3000°C. How long should the filament be in order to radiate 60 W of power? (σ = 5.67 * 10-8 W/m2 ∙ K4)

The filament should be 0.02935 m in order to radiate 60 W of power.

Explanation:

if A is the area of the filament, e is the emmisivity and T is the temperature of the filament then the power P is given by the stefan boltzmann power equation given by:

P = σ*A*e*T^4

A = P / (σ*e*T^4)

= (60) / [ (5.67*10^-8) * (1.0) * (3000 + 273) ^4]

= 9.22*10^-6 m^2

but if L is the length of the filament and d is the diameter of the filament then Area A is given by:

A = 2π*d*L

L = A / (2π*d)

= (9.22*10^-6) / (2π * (0.050*10^-3))

= 0.02935 m

Therefore, the filament should be 0.02935 m in order to radiate 60 W of power.