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Standard XII

Physics

Stefan-Boltzman's Law

A 100 watt fi...

Question

A 100 watt filament loses all its power by radiation, when it is heated to a temperature of 2500 K. If the diameter of the filament is 0.2 mm and the surface emissivity of the filament is 0.5, find the length of the filament [Stefan's constant $σ = 5.67 \times 10^{- 8} W / m^{2} K^{4}$ ]

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Solution

Given : $P = 100 W$ $T = 2500 K$ $e = 0.5$
Using $P = σ A e T^{4}$
where area of filament $A = π d l = π \times 0.2 \times 10^{- 3} l$
Or $100 = 5.67 \times 10^{- 8} \times (π) (0.2 \times 10^{- 3}) (l) \times 0.5 \times (2500)^{4}$
$⟹ l = 0.45 m$

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A 100 watt filament loses all its power by radiation, when it is heated to a temperature of 2500 K. If the diameter of the filament is 0.2 mm and the surface emissivity of the filament is 0.5, find the length of the filament [Stefan's constant σ=5.67×10−8 W/m2K4]

Given : P=100 W T=2500 K e=0.5Using P=σAeT4where area of filament A=πd l=π×0.2×10−3 lOr 100=5.67×10−8×(π)(0.2×10−3)(l)×0.5×(2500)4⟹ l=0.45 m

A 100 watt filament loses all its power by radiation, when it is heated to a temperature of 2500 K. If the diameter of the filament is 0.2 mm and the surface emissivity of the filament is 0.5, find the length of the filament [Stefan's constant $σ = 5.67 \times 10^{- 8} W / m^{2} K^{4}$ ]

Given : $P = 100 W$ $T = 2500 K$ $e = 0.5$
Using $P = σ A e T^{4}$
where area of filament $A = π d l = π \times 0.2 \times 10^{- 3} l$
Or $100 = 5.67 \times 10^{- 8} \times (π) (0.2 \times 10^{- 3}) (l) \times 0.5 \times (2500)^{4}$
$⟹ l = 0.45 m$