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28 April, 21:20

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)

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  1. 28 April, 21:39
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    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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