A sphere and a cube of same material and same total surface area are placed in the same evacuated space turn by turn after they are heated to the same temperature. Find the ratio of their initial rates of cooling in the enclosure.

Rate of emission of energy = σT4sLet m1 be the mass of sphere, C is specific heat and (dθdt), the rate of cooling.For sphereσT4S = m1C(dθdt)s------------(1)Let m2 be the mass of cube, C its specific heat and (dθdt), the rate of cooling.For cube σT4S = m2C(dθdt)c.----------(2)From equations (l) and (2)(dθdt)s(dθdt)c = m2 m1 = RsRcor    a3ρ(43)πr2ρ = RsRcwhere a is the side of cube and r is the radius of sphere, ρ is the density.∴ RsRc = 3a34πr3But since S is the same, so6a2 = 4πr2or       a2 = (23)πr2∴  RsRc = 3(2πr23)324πr3 = 2π2π3(4π)             = 2π12 = π6