Two black bodies P and Q have equal surface areas and are kept at temperatures 127^ C and 27^ C respectively. The ratio of thermal power radiated by A to that by B is

Step-by-step solution

The thermal power radiated by a black body is given by the Stefan-Boltzmann law: P = A T⁴ , where is the Stefan-Boltzmann constant, A is the surface area, and T is the absolute temperature in Kelvin. Given the temperatures in Celsius: T_ P = 127^ C = 127 + 273 = 400 K T_ Q = 27^ C = 27 + 273 = 300 K The ratio of thermal power radiated by P to that by Q is: P_ P P_ Q = A T_ P ⁴ A T_ Q ⁴ = ( T_ P T_ Q )⁴ Substituting the values: P_ P P_ Q = ( 400 300 )⁴ = ( 4 3 )⁴ Calculating the power: P_ P P_ Q = 4⁴ 3⁴ = 256 81 Ans