In an external environment of temperature ( T ) kelvin, a sphere at temperature ( 3T ) kelvin has cooling rate R₁ . When the temperature of that sphere falls to ( 2T ) kelvin, the cooling rate R₂ of the sphere will become

Step-by-step solution

According to Stefan-Boltzmann law, the rate of cooling of a body is given by R = dT dt = e A ms (T_ body ^4 - T_ surr ^4) . Let k = e A ms . For the first case, the temperature of the sphere is 3T and the surroundings is T . R₁ = k((3T)^4 - T^4) = k(81T^4 - T^4) = 80kT^4 For the second case, the temperature of the sphere falls to 2T . R₂ = k((2T)^4 - T^4) = k(16T^4 - T^4) = 15kT^4 Taking the ratio of R₂ to R₁ : R₂ R₁ = 15kT^4 80kT^4 = 15 80 = 3 16 R₂ = 3 16 R₁ Answer: 3 16 R₁