The temperature of perfect black body changes from 3000K to 4000K.What is the percentage change in energy radiation.

Asked by jatinsahu37 | 27th Mar, 2022, 04:45: PM

Expert Answer:

Radiation Power, P  = A σ T4
 
By taking logarithm ,  ln (P) = ln( A σ ) + 4 ln(T)
 
By differentiating both sides , we get , 
 
begin mathsize 14px style fraction numerator d P over denominator P end fraction space equals space 4 space fraction numerator d T over denominator T end fraction end style
if Temperature changes from 3000 K to 4000 K , then  ( dT/T ) = 1000/3000 = 0.333
 
Relative change in power , dP/P = 4 × 0.333 = 1.333
 
Percentage change in power  , (dP/P) × 100  = 133.3 %

Answered by Thiyagarajan K | 27th Mar, 2022, 05:23: PM