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The temperature of perfect black body changes from 3000K to 4000K.What is the percentage change in energy radiation.
Asked by jatinsahu37 | 27 Mar, 2022, 04:45: PM
Radiation Power, P  = A σ T4

By taking logarithm ,  ln (P) = ln( A σ ) + 4 ln(T)

By differentiating both sides , we get ,

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 %
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