A sphere is at temperature 600 K. In an external environment of 200 K, its cooling rate is R. When the temperature of the sphere falls to 400 K then cooling rate R' will become

Step-by-step solution

According to the Stefan-Boltzmann law, the rate of cooling of a body is given by R = dT dt = e A ms (T^4 - T₀^4) . For the first case, T = 600 K and T₀ = 200 K. R (600^4 - 200^4) R 200^4(3^4 - 1^4) = 200^4(81 - 1) = 80 200^4 For the second case, T = 400 K and T₀ = 200 K. R' (400^4 - 200^4) R' 200^4(2^4 - 1^4) = 200^4(16 - 1) = 15 200^4 Taking the ratio of the two rates: R' R = 15 200^4 80 200^4 = 15 80 = 3 16 R' = 3 16 R Answer: 3 16 R