A black rectangular surface of area A emits energy E per second at 127^ C. If length and breadth is reduced to half of initial value and temperature is raised to 527^ C then energy emitted becomes

Step-by-step solution

Using Stefan-Boltzmann law, the energy emitted per second by a black body is given by E = A T^4 . Initial temperature T₁ = 127^ C = 400 K Initial area A₁ = A Initial energy emitted E₁ = A (400)^4 = E Final temperature T₂ = 527^ C = 800 K Since length and breadth are reduced to half, the new area A₂ = L 2 B 2 = A 4 Final energy emitted E₂ = A₂ T₂^4 E₂ = ( A 4 ) (800)^4 E₂ = 1 4 16 A (400)^4 E₂ = 4 A (400)^4 = 4E Answer: 4E