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.

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