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 . Determine the radiation energy emitted per second by the ball A .

Step-by-step solution

According to the Stefan-Boltzmann law, the radiation energy emitted per second by a blackbody is given by: E = A T^4 For the spherical ball A , the given values are: A = 20 cm ^2 = 20 10⁻⁴ m ^2 T = 300 K 5.67 10⁻⁸ W m ⁻² K ⁻⁴ Substituting these values into the formula: E = (5.67 10⁻⁸) (20 10⁻⁴) (300)^4 E = 5.67 10⁻⁸ 20 10⁻⁴ 81 10^8 E = 5.67 1620 10⁻⁴ E 0.918 J/s The closest option provided is 0.94 J . (Note: Taking the Stefan-Boltzmann constant as = 5.8 10⁻⁸ W m ⁻² K ⁻⁴ yields exactly 0.9396 0.94 J/s ). Answer: 0.9