Question

A body cools in $7$ minutes from $60^\circ {\text{C}}$ to $40^\circ {\text{C}}$ . If the temperature of the surrounding is $10^\circ {\text{C}}$ , the temperature after next $7$ minutes will be:
A. $32^\circ {\text{C}}$
B. $38^\circ {\text{C}}$
C. $22^\circ {\text{C}}$
D. None of these

Answer 1

First of all, we will use the equation based on Newton’s law of cooling and then will integrate taking the boundaries over initial and final temperature. We will do the same for the second case too. We will compare two equations and manipulate accordingly.

Complete step by step answer:
First case,
In the given question, the following data are provided:
Initial temperature is $60^\circ {\text{C}}$ .
Final temperature is $40^\circ {\text{C}}$ .
Time required for cooling is $7$ minutes.

We will apply Newton’s law of cooling in this case, which is given by:
$\dfrac{{dT}}{{dt}} = k\left( {T - {T_s}} \right)$ …… (1)
Where,
$t$ indicates time required for cooling.
$T$ indicates temperature of the body.
${T_s}$ indicates temperature of the surrounding.

Now, we rearrange and perform integration in the equation (1) and we get:
\dfrac{{dT}}{{dt}} = k\left( {T - {T_s}} \right) \\\
\implies \int_{{T_i}}^{{T_f}} {\dfrac{{dT}}{{T - {T_s}}}} = \int_0^7 {kdt} \\\
\implies \left[ {\ln \left( {T - {T_s}} \right)} \right]_{{T_i}}^{{T_f}} = k\left[ t \right]_0^7 \\\
\implies \ln \left( {{T_f} - {T_s}} \right) - \ln \left( {{T_i} - {T_s}} \right) = k\left( {7 - 0} \right) \\\

Further simplifying the above expression, we get:
$\ln \left( {\dfrac{{{T_f} - {T_s}}}{{{T_i} - {T_s}}}} \right) = 7k$
$\dfrac{{{T_f} - {T_s}}}{{{T_i} - {T_s}}} = {e^{7k}}$ …… (2)

From the first case we substitute the required values in equation (2), we get:
\dfrac{{40 - 10}}{{60 - 10}} = {e^{7k}} \\\
\implies \dfrac{{30}}{{50}} = {e^{7k}} \\\
${e^{7k}} = \dfrac{3}{5}$ …… (3)
Second case,
Initial temperature will be$40^\circ {\text{C}}$ i.e. which was the final temperature of the first case.
We need to calculate the final temperature after an interval of $7$ minutes.
Again, we apply Newton’s law of cooling and we get from equation (2):
$\dfrac{{{T_f} - {T_s}}}{{{T_i} - {T_s}}} = {e^{7k}}$
$\dfrac{{{T_f} - 10}}{{40 - 10}} = {e^{7k}}$ …… (4)

Now, we compare equations (3) and (4):
\dfrac{{{T_f} - 10}}{{40 - 10}} = \dfrac{3}{5} \\\
\implies 90 = 5{T_f} - 50 \\\
\implies {T_f} = \dfrac{{140}}{5} \\\
\therefore {T_f} = 28^\circ {\text{C}} \\\

Hence, the temperature after next $7$ minutes will be $28^\circ {\text{C}}$ .

So, the correct answer is “Option D”.

Note:
While solving this problem, you will need some good knowledge in calculus. We use Newton’s law of cooling because this problem is based on heat loss which is directly proportional to the temperature difference between the body and its surroundings. It can also tell the speed of cooling of hot water present inside the pipes. For example, we can predict how long a cup of hot coffee will take to cool down to a desired temperature.