(a)
Area of the ball, A = 20 × 10−4 m2
Temperature of the ball, T = 57°C = 57 + 273 = 330 K
Amount of heat radiated per second = AσT4
= 20 × 10−4 × 6 × 10−8 × (330)4
= 1.42 J
(b)
Net rate of heat flow from the ball when its temperature is 200 is given by