The rate of dissipation of the heat by a black body at temperature T is Q . The rate of dissipation of heat by ordinary body of the same area at temperature 3T and having emissivity 0.25 , will be nearly

Step-by-step solution

According to Stefan-Boltzmann law, the rate of heat dissipation by a black body is given by Q = A T^4 . For an ordinary body with emissivity e and temperature T' , the rate of heat dissipation is Q' = e A T'^4 . Given e = 0.25 and T' = 3T , substituting these values gives: Q' = 0.25 A (3T)^4 Q' = 0.25 81 A T^4 Q' = 20.25 Q Therefore, the rate of dissipation is nearly 20 Q . Answer: 20 Q