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

9.3kviews

written 8.2 years ago by

teamques10 ★ 69k

• modified 4.9 years ago

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

heat transfer

ADD COMMENT EDIT

1 Answer

2.0kviews

written 8.2 years ago by

teamques10 ★ 69k

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℃

Create a free account to keep reading this post.

and 2 others joined a min ago.

ADD COMMENT EDIT

Please log in to add an answer.