A solid copper cube and sphere, both of same mass & emissivity are heated to the same initial temperature and kept under identical conditions. What is the ratio of their initial rate of fall of temperature?

Hint:
- To solve this question, we will use the law of Newton’s law of cooling which is used to determine the ratio of the initial rate of fall of temperature. Then using the formulas of volume and surface areas of cube and sphere, we will get our required answer.
- Formula Used:
$
  1.Density = \dfrac{{Mass}}{{Volume}} \\
  2.Volume{\text{ of cube = (side}}{{\text{)}}^3} \\
  3.Volume{\text{ of }}sphere = \dfrac{4}{3}\pi {(radius)^3} \\
  4.Suface{\text{ }}area{\text{ }}of{\text{ }}cube = 6{(side)^2} \\
  5.Suface{\text{ }}area{\text{ }}of{\text{ }}sphere = 4\pi {(side)^2} \\
$

Complete step by step solution:
We have been given that a solid copper cube and sphere, both have the same mass & emissivity and are heated to the same initial temperature and kept under identical conditions. We need to find the ratio of their initial rate of fall of temperature.

To determine the rate of cooling or rate of drop of temperature, Newton’s law of cooling is used; which is,
Newton’s law of cooling $ = \dfrac{{dT}}{{dt}}[ - KA(T - {T_0})]$\[ \ldots ..eq.\left( 1 \right)\]

Now, since both cube and sphere are made up of the same material and have the same mass, so their density will be equal.
Let ‘a’ be the length of the cube and let ‘r’ be the radius of the sphere.

So, the density of cube \[ = \] density of sphere
$ \Rightarrow \dfrac{{{\text{mass }}of{\text{ cube}}}}{{{\text{volume }}of{\text{ cube}}}} = \dfrac{{{\text{mass }}of{\text{ }}sphere}}{{{\text{volume }}of{\text{ }}sphere}}.....................(\because density = \dfrac{{mass}}{{volume}})$

Since, density is equal, mass is equal, so the volume of the cube is also equal to the volume of the sphere.
\[
   \Rightarrow {a^3} = \dfrac{4}{3}\pi {r^3} \\
  \dfrac{a}{r} = {(\dfrac{{4\pi }}{3})^{\dfrac{1}{3}}}.........eq.(2) \\
\]

Now, from newton’s law of cooling mentioned above, K will be equal for both cube and sphere, since K depends upon material and emissivity, \[\left( {T - {T_0}} \right)\] will be equal, since both having same initial temperature and kept under identical conditions. So, $\dfrac{{dT}}{{dt}}$ will only depend upon area.

\[\dfrac{{{{(dT/dt)}_{cube}}}}{{{{(dT/dt)}_{sphere}}}} = \dfrac{{Are{a_{cube}}}}{{Are{a_{sphere}}}} = \dfrac{{6{a^2}}}{{4\pi {r^2}}} = \dfrac{6}{{4\pi }}{(\dfrac{a}{r})^2}\]

From eq. 2, we get
\[\dfrac{{Are{a_{cube}}}}{{Are{a_{sphere}}}} = \dfrac{6}{{4\pi }}{(\dfrac{{4\pi }}{3})^{\dfrac{2}{3}}} = {(\dfrac{6}{\pi })^{^{\dfrac{1}{3}}}}\]

So, the ratio of their initial rate of fall of temperature is \[{(\dfrac{6}{\pi })^{\dfrac{1}{3}}}.\]

Note:
In the solutions, we have mentioned Newton's law of cooling, let us understand about that in detail. This law states that the rate of heat loss of a body is directly proportional to the temperature difference between the body and its surroundings, and this law is used to determine the fall of temperature.