Question

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 rates of cooling in the enclosure:
$(A)\sqrt {\dfrac{\pi }{6}} :1$
$(B)\sqrt {\dfrac{\pi }{3}} :1$
$(C)\dfrac{\pi }{{\sqrt 6 }}:1$
$(D)\dfrac{\pi }{{\sqrt 3 }}:1:1$

Answer 1

In order to find the ratio of the rate of cooling of the sphere and the cube, we will first find the ratio of the rate of emission of energy and with the help of this ratio, we will find a relation of it with the rate of cooling of the two bodies.

Complete step by step solution:
We know that the rate of emission of energy is given by the expression,
Rate of emission of energy $= \sigma {T^4}s$
According to the question,
Let us consider the mass of the sphere as ${m_2}$, also $C$ is the specific heat and let $\dfrac{{d\theta }}{{dt}}$ be the rate of cooling of the sphere. So, for the sphere,
$\sigma {T^4}S = {m_1}C{\left( {\dfrac{{d\theta }}{{dt}}} \right)_s}.....(1)$
Similarly, Let us consider the mass of the cube as ${m_1}$, also $C$ is the specific heat and let $\dfrac{{d\theta }}{{dt}}$ be the rate of cooling of the cube. So, for the cube,
$\sigma {T^2}S = {m_2}C{\left( {\dfrac{{d\theta }}{{dt}}} \right)_c}.....(2)$
From equation (1) and (2), we can say that,
$\dfrac{{{{\left( {d\theta /dt} \right)}_s}}}{{{{\left( {d\theta /dt} \right)}_c}}} = \dfrac{{{m_2}}}{{{m_1}}} = \dfrac{{{R_s}}}{{{R_c}}}$
Or, we can say that,
$\dfrac{{{a^3}\rho }}{{\left( {4/3} \right)\pi {r^2}\rho }} = \dfrac{{{R_s}}}{{{R_c}}}$
In the above equation,
$a$ is the side of the cube
$r$ is the radius of the sphere
$\rho$ is the density
On further simplifying the above equation, we get
$\dfrac{{{R_s}}}{{{R_c}}} = \dfrac{{3{a^3}}}{{4\pi {r^3}}}$
But as we know that $S$ is the same, so,
$6{a^2} = 4\pi {r^2}$
This can be written as,
${a^2} = \dfrac{2}{3}\pi {r^2}$
So, $\dfrac{{{R_s}}}{{{R_c}}} = \dfrac{{3{{\left( {2\pi {r^2}/3} \right)}^{3/2}}}}{{4\pi {r^3}}}$
On further solving,
$\dfrac{{{R_s}}}{{{R_c}}} = \dfrac{{2\pi \sqrt {2\pi } }}{{\sqrt 3 (4x)}}$
So, we can write,
$\sqrt {\dfrac{{2x}}{{12}}} = \sqrt {\dfrac{\pi }{6}}$
So, the final answer is $(A)\sqrt {\dfrac{\pi }{6}} :1$

Note:
In accordance with Newton's law of cooling, the time rate of loss of heat from a body is always directly proportional to the difference in the temperature of the given body and the surroundings of the body. But the biggest limitation of this law is that the difference in temperature between the body and its surroundings must be small for this law to be applicable.