Question

A body cools from $70^\circ {\text{C}}$ to $60^\circ {\text{C}}$ in 8 minutes. The same body cools from $60^\circ {\text{C}}$ to $50^\circ {\text{C}}$ in
A) $8{\text{ minute}}$
B) ${\text{less than }}8{\text{ minute}}$
C) ${\text{more than }}8{\text{ minute}}$
D) $1{\text{ or 2 or 3 depending on specific heat of the body}}$

Answer 1

Here, we will be using Newton’s law of cooling. It states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. First of all, we will calculate the constant used in Newton’s law of cooling from the initial given condition. Using that, we will calculate the time required considering the given conditions.

Complete step by step answer:
In this question, a body is cooling from $70^\circ {\text{C}}$ to $60^\circ {\text{C}}$ in 8 minute and we have to calculate the time required to cool the same body from $60^\circ {\text{C}}$ to $50^\circ {\text{C}}$.
Frist, write the expression for Newton’s law of cooling as follows:
$\dfrac{{d\theta }}{{dt}} = - k\left( {{\theta _{avg}} - {\theta _0}} \right)$ …… (1)
Here, ${\theta _0}$ is the surrounding temperature and ${\theta _{avg}}$ is the average of two temperatures.
Let’s assume surrounding temperature is $25^\circ {\text{C}}$.
Apply condition 1 to the equation (1) as follows,
$\Rightarrow \dfrac{{d\theta }}{{dt}} = - k\left( {{\theta _{avg}} - {\theta _0}} \right)$
$\Rightarrow \dfrac{{70 - 60}}{8} = - k\left( {\dfrac{{70 + 60}}{2} - 25} \right)$
$\Rightarrow \dfrac{{10}}{8} = - k\left( {65 - 25} \right)$
$\Rightarrow k = - \dfrac{{10}}{{320}}$ …………….…… (2)
Now, apply condition 2 to the equation (1) as follows:
$\Rightarrow \dfrac{{d\theta }}{{dt}} = - k\left( {{\theta _{avg}} - {\theta _0}} \right)$
$\Rightarrow \dfrac{{60 - 50}}{t} = - \left( { - \dfrac{{10}}{{320}}} \right)\left( {\dfrac{{60 + 50}}{2} - 25} \right)$
$\Rightarrow \dfrac{{10}}{t} = \dfrac{{10}}{{320}}\left( {55 - 25} \right)$
Simplify further to obtain the time required as
$\Rightarrow \dfrac{{10}}{t} = \dfrac{{10}}{{320}} \times 30$
$\Rightarrow t = \dfrac{{320 \times 10}}{{10 \times 30}}$
$\Rightarrow t = 10.67{\text{ minute}}$

So, from the above calculation, the time required to cool the same body from $60^\circ {\text{C}}$ to $50^\circ {\text{C}}$ is $10.67{\text{ minute}}$ and it is more than 8 minutes. Therefore, the correct option is C.

Note:
Observe the negative sign before constant k. It represents that the rate of cooling will decrease with a decrease in temperature. Because of this, we can say the time will increase. While applying Newton’s law of cooling, one should use the average temperature in the expression. If someone uses different surrounding temperatures then also, he will get the same results.