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A filament of 75 W Light bulb may be considered as black body radiating into a black enclosure of 70deg C

written 7.9 years ago by teamques10 ★ 68k

•   modified 4.5 years ago

The filament diameter is 0.1 mm and length 50 mm. considering the radiation, determine the filament temperature.

heat transfer

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written 7.9 years ago by teamques10 ★ 68k

Q=75 W= 75 J/s

T_2=70+273=343

d=0.1 mm

L=5 cm

Area=πdl

ϵ=1 for black body

Q=σ∈A($T_1^4-T_2^4$ ) 75

=5.67×$10^(-8)×π×0.1×10^(-3)×5×10^(-2)$ ($T_1^4-343^4$)

$T_1$=3029-273=2756℃

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