A cube and a sphere of equal edge and radius, made of the same substance are allowed to cool under identical conditions. Determine which of the two will cool at a faster rate.

Explanation

From equation, the rate of cooling of a body is given by
`(-dT)/(dt)=(Aesigma)/(rhoVs)(T_4-T_0^4)`
since, substance is same for both bodies, so
`e//rho s=`constant.
Finally, thay are allowed to cool under identical conditions so `(T^4-T_0^4)=` constant.
`(-dT)/(dt)prop(A)/(V)`
Let the edge of the cube or radius of the sphere be a, then for cube, `A=6a^2` and `V=a^3` so `(A)/(V)=6a`
For the sphere, `A=4pia^2` and `V=(4//3)pia^3`, so `A//V=3//a`
Evidently, the ratio `A//V` is more for cube, so the cube cools at a faster rate. Not the special technique used in this problem.