Question

A body cools down from ${80^ \circ }C$ to ${60^ \circ }C$ in 10 minutes when the temperature of surrounding is ${30^ \circ }C$ . The temperature of the body after next 10 minutes will be:
A) ${30^ \circ }C$
B) ${48^ \circ }C$
C) ${50^ \circ }C$
D) ${52^ \circ }C$

Answer 1

Hint: We can find the answer to this question by using Newton's law of cooling. By using the equation for Newton's law of cooling for the first 10 minute we can get the value of the constant in the equation and using that and the other given values we can find the temperature after the next 10 minute.

Complete step by step solution:
It is given that a body cools from ${80^ \circ }C$ to ${60^ \circ }C$ in 10 minutes.
Let us denote the initial temperature as ${\theta _1}$ and final temperature as ${\theta _2}$
Let the time taken be t
The surrounding temperature is denoted as ${\theta _0}$ .
We can use Newton’s law of cooling in this case. Newton’s law of cooling states that the rate of loss of heat is directly proportional to the temperature difference between the body and the surrounding.
According to Newton's law of cooling we have the equation
$\dfrac{{{\theta _1} - {\theta _2}}}{t} = \alpha \left[ {\dfrac{{{\theta _2} + {\theta _1}}}{2} - {\theta _0}} \right]$
Where ${\theta _1}$ and ${\theta _2}$ is initial and final temperature t is the time taken alpha is a constant. ${\theta _0}$ is a surrounding temperature.
$\dfrac{{{\theta _2} + {\theta _1}}}{2}$ denotes the average temperature.
Now let us substitute the given values for the first case. In the first ten seconds we have
${\theta _1} = {80^ \circ }$
${\theta _2} = {60^ \circ }$
Time , $t = 10\min $
surrounding temperature, ${\theta _0} = {30^ \circ }$
On substituting these values in the newton's law of cooling equation, we get,
$\dfrac{{{{80}^ \circ } - {{60}^ \circ }}}{{10}} = \alpha \left[ {\dfrac{{{{80}^ \circ } + {{60}^ \circ }}}{2} - {{30}^ \circ }} \right]$
$\dfrac{{{{20}^ \circ }}}{{10}} = \alpha \left[ {{{40}^ \circ }} \right]$
From this we get the value of alpha as
$\alpha = \dfrac{2}{{40}}$
Now let's take the second case. That is for the next 10 minutes.
Now the initial temperature is ${60^ \circ }$
Let the final temperature be denoted as $\theta $
The temperature of the surrounding is the same therefore ${\theta _0} = {30^ \circ }$
Now let us substitute all these values to find the value of final temperature
$\dfrac{{{{60}^ \circ } - \theta }}{{10}} = \dfrac{2}{{40}}\left[ {\dfrac{{{{60}^ \circ } + \theta }}{2} - {{30}^ \circ }} \right]$
$ \Rightarrow {60^ \circ } - \theta = \dfrac{1}{2}\left[ {{{\dfrac{{{{60}^ \circ } + \theta - 60}}{2}}^ \circ }} \right]$
$ \Rightarrow {60^ \circ } - \theta = \dfrac{\theta }{4}$
$\therefore \theta = {48^ \circ }$
This is the temperature of the body after the next 10 minutes.

Thus, the correct answer is option B.

Note: Newton's law of cooling cannot be applied if there is a large difference in temperature between the body and the surrounding. Also when the temperature of the surroundings keeps changing, that is if ${\theta _0}$ is not a constant , then newton's law of cooling is not valid.