Question

A solid sphere cools at the rate of $2.8{}^{\circ }C/\min $, when its temperature is $127{}^{\circ }C$. The rate at which another solid sphere of the same material of twice the radius will lose its temperature at $327{}^{\circ }C$ is given by (take the surrounding temperature at $27{}^{\circ }C$ ).
A. $6.8{}^{\circ }C/\min $
B. $5.6{}^{\circ }C/\min $
C. $4.2{}^{\circ }C/\min $
D. $8.4{}^{\circ }C/\min $

Answer 1

Hint: Use the formula for the art of cooling given by Newton’s law of cooling. Write the equation of rate of cooling for the solid spheres in both cases. Then after some mathematical operations calculate the rate of cooling of the larger sphere.

Formula used:
$\dfrac{dT}{dt}=\dfrac{kA}{m}(T-{{T}_{0}})$
where $\dfrac{dT}{dt}$ is the rate of cooling of a body with surface area A and mass m at temperature at T. k is a constant that depends on the material of the body and ${{T}_{0}}$ is the temperature of the surrounding.
$m=\rho V$
 where m is mass, $\rho $ is density and V is volume of the body.

Complete step by step answer:
Here, the value of k and density ($\rho $) is the same for both the solid spheres as it is said that both are of the same material. In the first case, the rate of cooling of the solid sphere is $\dfrac{dT}{dt}=\dfrac{k{{A}_{1}}}{{{m}_{1}}}({{T}_{1}}-{{T}_{0}})$ ….. (i)
In this case, let the radius of the sphere be ${{r}_{1}}$.
Then the surface area of the sphere is ${{A}_{1}}=\pi r_{1}^{2}$.
The mass of this sphere is ${{m}_{1}}=\rho {{V}_{1}}$
And ${{V}_{1}}=\dfrac{4}{3}\pi r_{1}^{3}$
Then,
$\Rightarrow {{m}_{1}}=\rho \dfrac{4}{3}\pi r_{1}^{3}$
Now, substitute the known values in equation (i).
$\Rightarrow \dfrac{dT}{dt}=\dfrac{k\pi r_{1}^{2}}{\rho \dfrac{4}{3}\pi r_{1}^{3}}({{T}_{1}}-{{T}_{0}})$
$\Rightarrow \dfrac{dT}{dt}=\dfrac{3k}{\rho 4{{r}_{1}}}({{T}_{1}}-{{T}_{0}})$ …. (ii)
But it is given that for this sphere $\dfrac{dT}{dt}=2.8{}^{\circ }C/\min $ and ${{T}_{1}}=127{}^{\circ }C,{{T}_{0}}=27{}^{\circ }C$.
Substitute these values in equation (ii).
$\Rightarrow 2.8=\dfrac{3k}{\rho 4{{r}_{1}}}(127-27)$
$\Rightarrow 2.8=\dfrac{300k}{\rho 4{{r}_{1}}}$ ….. (iii).
In the second case, the rate of cooling of the solid sphere is $\dfrac{dT}{dt}=\dfrac{k{{A}_{2}}}{{{m}_{2}}}({{T}_{2}}-{{T}_{0}})$ ….. (iv)
In this case, let the radius of the sphere be ${{r}_{2}}$.
Then the surface area of the sphere is ${{A}_{2}}=\pi r_{2}^{2}$.
The mass of this sphere is ${{m}_{2}}=\rho {{V}_{2}}$
And ${{V}_{2}}=\dfrac{4}{3}\pi r_{2}^{3}$
Then,
${{m}_{2}}=\rho \dfrac{4}{3}\pi r_{2}^{3}$
Now, substitute the known values in equation (iv).
$\Rightarrow \dfrac{dT}{dt}=\dfrac{k\pi r_{2}^{2}}{\rho \dfrac{4}{3}\pi r_{2}^{3}}({{T}_{2}}-{{T}_{0}})$
$\Rightarrow \dfrac{dT}{dt}=\dfrac{3k}{4\rho {{r}_{2}}}({{T}_{2}}-{{T}_{0}})$ …. (v)
But it is given that for this sphere ${{T}_{2}}=327{}^{\circ }C,{{T}_{0}}=27{}^{\circ }C$.
Substitute these values in equation (ii).
$\dfrac{dT}{dt}=\dfrac{3k}{4\rho {{r}_{2}}}(327-27)$
$\Rightarrow \dfrac{dT}{dt}=\dfrac{900k}{4\rho {{r}_{1}}}$ ….. (vi).
Now, divide (vi) by (iii).
$\dfrac{\dfrac{dT}{dt}}{2.8}=\dfrac{\dfrac{900k}{4\rho {{r}_{2}}}}{\dfrac{300k}{4\rho {{r}_{1}}}}$
$\Rightarrow \dfrac{dT}{dt}=2.8\times \dfrac{900{{r}_{1}}}{300{{r}_{2}}}$
But it is given that $\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{1}{2}$.
$\Rightarrow \dfrac{dT}{dt}=2.8\times \dfrac{900{{r}_{1}}}{300{{r}_{2}}}\\
\Rightarrow \dfrac{dT}{dt} = 2.8\times \dfrac{900}{300}\times \dfrac{1}{2}\\
\therefore \dfrac{dT}{dt} = 4.2{}^{\circ }C/\min $
This means that the rate of cooling of the second sphere is $4.2{}^{\circ }C/\min $.

Hence, the correct option is C.

Note:The formula for the rate of cooling of a body given by Newton’s law of cooling only works when the difference between the temperature of the body and the temperature of the surrounding is very less. If this temperature difference is large then Newton's law of cooling is invalid.