Answer:
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
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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