Question

A filament maintained at temperature 1827 degree Celsius has emissivity 0.5 . calculate power radiated per unit area of the filament.​

Answer 1

Answer:

plzzzzzzz change the value you get your answer

Explanation:

T=2000K, A=5 x 10

−5

m

2

,ε=0.85σ=5.7×10

−8

MKS unit.

Energy radiated/sec = AεσT

4

=5×10

−5

×0.85×5.7×10

−8

x(2000)

4

=5×0.85×537×10

−13

×10

12

×16

=387.6×10

−1

Energy radiated/minute =38.76×60

= 2325.6J