11. A 100-W incandescent light bulb has a cylindrical tungsten filament 30.0 cm long, 0.40 mm in diameter, and with an emissivity of 0.26. (a) What is the temperature of the filament

The temperature of the filament is 366.3 K

Using the equation for radiated power, P

P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²K⁴, ε = emissivity = 0.26, A = area of cylindrical tungsten filament = πDh (since it radiates through its sides)where D = diameter of tungsten filament = 0.40 mm = 0.4 × 10⁻³ m and h = length of tungsten filament = 30.0 cm = 0.3 m and T = temperature of tungsten filament.

Since we require the temperature of the filament, making T subject of the formula, we have

T = ⁴√(P/σεA)

T = ⁴√(P/σεπDh)

Since P = 100 W, substituting the values of the variables into the equation, we have

T = ⁴√(P/σεπDh)

T = ⁴√(100 W/5.67 × 10⁻⁸ W/m²K⁴× 0.26 × π× 0.4 × 10⁻³ m × 0.3 m)

T = ⁴√(100 W/0.5558 W/K⁴ × 10⁻⁸)

T = ⁴√179.93 × 10⁸ K⁴)

T = 3.663 × 10² K

T = 366.3 K

So, the temperature of the filament is 366.3 K

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 T = 556K

Explanation:

For this exercise let's use Stefan's law

        P = σ A e T⁴

where the stafan-Boltzmann constant is 5,670 10⁻⁸ W / m² K⁴.

In this case the area of a circular cylinder is

        A = π r²

the radius is

         r = d / 2 = 0.150 m

        A = π (0,150)²

        A = 7.0685 10⁻² m²

let's clear

         T⁴ = P /σ A e

         T⁴ = 100 / (5.670 10⁻⁸   7.0685 10⁻²  0.26)

         T⁴ = 9.5966 10¹⁰ = 959.66 10⁸

         T = 5.56 102 K

         T = 556K

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explanation: