Question

An ordinary body A and a perfect black body B are maintained at the same temperature. If the radiant power of A is 2x104 W/m2 and that of B is 5x104 W/m2 then the coefficient of emission of A is
​

Answer 1

Answer:

1701 J

Explanation:

The power radiated is given by Stefan's law:

P=σAeT  

4

 

Stefan-Boltzmann constant: σ=5.67×10  

−8

Wm  

−2

K  

−4

 

A=5×10  

−4

m  

2

 

Temperature, T =727  

∘

C=1000K

e=1 for a black body

P=5.67×10  

−8

×5×10  

−4

×1×(1000)  

4

=28.35W

Heat lost in one minute =60×28.35=1701J