Question

Two metallic spheres $S_{1}$ and $S_{2}$ are made of the same material and have identical surface finish. The mass of $S_{1}$ is thrice that of $S_{2}$. Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of $S_{1}$ to that of $S_{2}$ is\\
A. $\dfrac{1}{3}$
B. $\dfrac{3}{1}$
C. $\sqrt{3}$
D. $\dfrac{\sqrt{3}}{1}$
E. $\left(\dfrac{1}{3}\right)^{\frac{1}{3}}$

Answer 1

Hint: We have to find the initial rate of cooling of the first sphere with respect to the second sphere. We can solve this question by combining the equation for heat energy radiating through the surface and equation for heat energy in terms of rate of heat flow through the sphere.






Formula used:
There are two formulae used
(1) Rate of energy radiated through the object is:
$H=A \sigma T^{4}$
Where $\mathrm{A}$ is the area (Here area of the sphere. That is $A=4 \pi \mathrm{r}^{2}$, where $\mathrm{r}$ is the radius of the sphere.), $\mathrm{T}$ is temperature and $\sigma$ is the Steffan-Boltzmann constant.
(2) Rate of heat, $H=\dfrac{d Q}{d t}=m s \dfrac{d \theta}{d t}$
Where $\mathrm{m}$ is the mass of the sphere and $\mathrm{s}$ is the specific heat capacity and $d \theta$ is the temperature gradient.





Complete answer:
We have two metallic spheres. They are made with the same material. The mass of sphere $S_{1}$ is thrice that of $S_{2}$. They are heated to the same temperature and cool down individually. We have to find the initial rate of cooling of $S_{1}$ to that of $S_{2}$.
We have two equations for the rate of heat.
(1) Rate of energy radiated through the object is:
$H=A \sigma T^{4}$
Where $\mathrm{A}$ is the area (Here area of the sphere. That is $A=4 \pi \mathrm{r}^{2}$, where $\mathrm{r}$ is the radius of the sphere.), $\mathrm{T}$ is temperature and $\sigma$ is the Steffan-Boltzmann constant.

(2) Rate of heat, $H=\dfrac{d Q}{d t}=m s \dfrac{d \theta}{d t}$
Where $\mathrm{m}$ is the mass of the sphere and $\mathrm{s}$ is the specific heat capacity and $d \theta$ is the temperature gradient.
We can equate it to find the relation connecting rate of cooling and mass of the sphere since mass is the only given quantity.


On equating both equations, we get:
$m s \dfrac{d \theta}{d t}=A \sigma T^{4}$
We can simplify this equation again as:
$$
\dfrac{d \theta}{d t}=\dfrac{A \sigma T^{4}}{m s}=\dfrac{\sigma 4 \pi r^{2} T^{4}}{m s}=K\left(\dfrac{r^{2}}{m}\right)-(1)
$$
Where $K=\dfrac{\sigma 4 \pi T^{4}}{s}$ which is a constant.
Therefore, rate of cooling is directly proportional to square of radius and inversely proportional to mass of the sphere.
Now we know the relation between mass, volume and radius of the sphere.
Density, $\rho=\dfrac{\text { mass }}{\text { volume }}=\dfrac{m}{\dfrac{4}{3} \pi \mathrm{r}^{3}}$
We have relation for radius and mass
$r \propto m^{\dfrac{1}{3}}-(2)$
Equating equation (1) and (2) we get:
Rate of cooling, $\dfrac{d \theta}{d t} \propto \dfrac{m^{\dfrac{2}{3}}}{m}$
That is, $\dfrac{d \theta}{d t} \propto \dfrac{1}{m^{\dfrac{1}{3}}}$
Now comparing rate of cooling of both spheres we get,
We have equation for rate of cooling for first sphere, $\left(\dfrac{d \theta_{1}}{d t}\right)=\left(\dfrac{1}{m_{1}^{\dfrac{1}{3}}}\right)$
We have equation for rate of cooling for second sphere, $\left(\dfrac{d \theta_{2}}{d t}\right)=\left(\dfrac{1}{m_{2}^{\dfrac{1}{3}}}\right)$
Therefore, initial rate of cooling of $S_{1}$ to that of $S_{2}$ is:
$$
\dfrac{d \theta_{1}}{d \theta_{2}}=\left(\dfrac{m_{2}}{m_{1}}\right)^{\dfrac{1}{3}}
$$
It is given that $m_{1}=3 m_{2}$
Therefore, initial rate of cooling of $S_{1}$ to that of $S_{2}$ is:
$$
\dfrac{d \theta_{1}}{d \theta_{2}}=\left(\dfrac{1}{3}\right)^{\dfrac{1}{3}}
$$
Therefore, the answer is option (D)




Note:
Note that here the rate of cooling is occurring at the same atmosphere but they are insulated from each other and both spheres have the same material that is why all other quantities remain constant.