A hollow cube `A` of very small thickness at temperature `T` emits radiation like a black body. The cube `A` is kept in a room of temperatue `T_(0)(ltT)`. The side of cube is doubled without changing the volume of material and its temperature. Find the ratio of rate of cooling of cube `A` in first case to that of second case.

Answer

4

Explanation

The rate of heat loss by cube of thickness `Deltax`, side `l` and made of density `rho` is given by
`-md(dT)/(dt)=epsilonsigmaA(T^(4)-T_(0)^(4))`
`-(dT)/(dt)=(epsilon sigma A(T^(4)-T_(0)^(4)))/(ms)`