The rectangular surface area (12 cm 6 cm ) of the black body, at a temperature 127^ C , emits heat energy at the rate of Q units per second. If the length and breadth of the surface area are each reduced to half of its initial value and the temperature is increased to 327^ C , then the emission of heat energy will be

Step-by-step solution

According to Stefan-Boltzmann law, the rate of heat emission is given by E = A T^4 . Initial area A₁ = 12 6 = 72 cm ^2 Initial temperature T₁ = 127^ C = 400 K Initial rate of heat emission Q₁ = Q Final area A₂ = 12 2 6 2 = 18 cm ^2 = A₁ 4 Final temperature T₂ = 327^ C = 600 K Using the relation Q A T^4 , we get: Q₂ Q₁ = A₂ A₁ ( T₂ T₁ )^4 Q₂ Q = 1 4 ( 600 400 )^4 Q₂ Q = 1 4 ( 3 2 )^4 = 1 4 81 16 = 81 64 Q₂ = 81Q 64 Answer: 81Q 64