A spherical ball A having a surface area of 20 cm ^2 is positioned at the centre of a hollow spherical shell B of area 80 cm ^2 . The surface of A and the inner surface of B both emit as blackbodies. A and B are both maintained at a temperature of 300 K . Calculate the radiation energy emitted per second by the inner surface of B .

Step-by-step solution

According to the Stefan-Boltzmann law, the total radiation energy emitted per second by a blackbody surface is given by: E = A T^4 For the inner surface of the hollow spherical shell B : Area, A_B = 80 cm ^2 = 80 10⁻⁴ m ^2 Temperature, T = 300 K Stefan-Boltzmann constant, 5.67 10⁻⁸ W m ⁻² K ⁻⁴ Substituting the given values into the formula: E_B = 5.67 10⁻⁸ 80 10⁻⁴ (300)^4 E_B = 5.67 10⁻⁸ 80 10⁻⁴ 81 10^8 E_B = 5.67 80 81 10⁻⁴ E_B = 3.674 J/s Rounding to the nearest given option, we get approximately 3.8 J . (Note: I