Question

A body cools down from $60^\circ {\rm{C}}$ to $55^\circ {\rm{C}}$ in $5$ minutes. It will cool down from $55^\circ C$ to $50^\circ {\rm{C}}$ in
(A) More than $5$ minutes
(B) Can’t be predicted
(C) $5$ minutes
(D) Less than $5$ minutes

Answer 1

Hint:
This question is based on Newton’s Law of Cooling. According to this law, “The rate of decrease in the temperature of the body is directly related to the difference in the temperature between the body and the surroundings.”
The formula for the rate of decrease in the temperature of the body is given by $\dfrac{{dT}}{{dt}} = - hA\left( {T - {T_s}} \right)$
Where, $\dfrac{{dT}}{{dt}}$ is the decrease in the temperature of the body with respect to time
$h$ is the heat transfer coefficient of the body
$A$ is the surface area of the body, and
$\left( {T - {T_s}} \right)$ is the temperature difference between body and surroundings.

Complete step by step solution :
Let us assume the Surrounding temperature be ${T_s}$
The temperatures for the first case,
${T_1} = 60^\circ C$ and ${T_2} = 55^\circ C$
The temperatures for the second case,
${T_2} = 55^\circ C$ and ${T_3} = 50^\circ C$
The average temperature in the first case
 $
{T_{avg}} = \dfrac{{{T_1} + {T_2}}}{2}\\
{\Rightarrow T_{avg}} = \dfrac{{60 + 55}}{2}\\
{\Rightarrow T_{avg}} = 57.5^\circ {\rm{C}}
$
And the difference between the average temperature for the first case and the surroundings $\left( {{T_{avg}} - {T_s}} \right) = \left( {57.5 - {T_s}} \right)$
The change in temperature with respect to time for the first case is given by,
$
\dfrac{{dT}}{{dt}} = \dfrac{{\left( {60 - 55} \right)^\circ C}}{{5{\rm{ min}}}}\\
 = \dfrac{5}{5}\\
 = 1^\circ {\rm{C \ \ per \, min}}
$

By applying Newton’s law of cooling, we have.

$
\dfrac{{dT}}{{dt}} = - hA\left( {{T_{avg}} - {T_s}} \right)\\
\Rightarrow 1 = - hA\left( {57.5 - {T_s}} \right)
$
So,
$ \Rightarrow - hA = \dfrac{1}{{\left( {57.5 - {T_s}} \right)}}{\rm{ mi}}{{\rm{n}}^{ - 1}}$
Now the average temperature for second case
$
{{T'}_{avg}} = \dfrac{{{T_2} + {T_3}}}{2}\\
= \dfrac{{55 + 50}}{2}\\
= 52.5^\circ {\rm{C}}
$
And the difference between the average temperature for the second case and the surroundings $\left( {{{T'}_{avg}} - {T_s}} \right) = \left( {52.5 - {T_s}} \right)$
The change in temperature with respect to time for the second case is given by,
$
\dfrac{{dT}}{{dt}} = \dfrac{{\left( {55 - 50} \right)^\circ C}}{{t{\rm{ min}}}}\\
 = \dfrac{5}{t}{\rm{^\circ C \ \ per \, min}}
$
By applying Newton’s law of cooling, we have-

$
\dfrac{{dT}}{{dt}} = - hA\left( {{{T'}_{avg}} - {T_s}} \right)\\
\Rightarrow \dfrac{5}{t} = - hA\left( {52.5 - {T_s}} \right)
$
Substituting the value $ - hA = \dfrac{1}{{\left( {57.5 - {T_s}} \right)}}{\rm{ mi}}{{\rm{n}}^{ - 1}}$ in this equation we get,
$\dfrac{5}{t} = \dfrac{1}{{\left( {57.5 - {T_s}} \right)}} \times \left( {52.5 - {T_s}} \right)$
Solving for $t$ we have-

$t = \dfrac{{5 \times \left( {57.5 - {T_s}} \right)}}{{\left( {52.5 - {T_s}} \right)}}{\rm{ min}}$
Now if we solve this equation, we know that the answer would be greater than $5$.
So, the answer would be $t > {\rm{5 minutes}}$.

Therefore, the correct option is (A).

Note:
Newton’s law of cooling is applicable only in the case of fluids (gases and liquids), not in solids. For solids, the Fourier law is applied to find the heat transfer in the body.