A black rectangular surface of area 'A' emits energy 'E' per second at 27^ C. If length and breadth is reduced to half of its initial value and temperature is raised to 327^ C then energy emitted per second becomes

Step-by-step solution

Initial temperature T₁ = 27^ C = 300 K Initial area A₁ = A Energy emitted per second E = A T₁^4 = A (300)^4 When length and breadth are reduced to half, the new area A₂ = L 2 B 2 = A 4 New temperature T₂ = 327^ C = 600 K New energy emitted per second E' = A₂ T₂^4 E' = ( A 4 ) (600)^4 E' = ( A 4 ) (2 300)^4 E' = ( A 4 ) 16 (300)^4 E' = 4 A (300)^4 = 4E Answer: 4E