Question

The temperature of a body in a room is $100{}^\circ F$ . After five minutes, the temperature of the body becomes $50{}^\circ F$ . After another 5 minutes, the temperature becomes $40{}^\circ F$ . What is the temperature of surroundings?

Answer 1

Hint: We should know Newton's Cooling Law which is given as $\dfrac{dt}{dT}\propto \left( T-S \right)$ . After this, we have to do integration on both sides and the equation will be getting as $\log \left( T-S \right)=-kt+c$ . Then we have to find the constant term c by putting T as $100{}^\circ F$ at $t=1$ . Then on getting the final equation we will put values given in question and then will make k as subject variable and equate both the equation. From this, we will get the surrounding temperature.

Complete step-by-step answer:
Here, we will use the formula according to Newton's Cooling Law given as $\dfrac{dt}{dT}\propto \left( T-S \right)$ assuming at time t and temperature of a body as T.
As the temperature of body is decreasing continuously, we will assume proportionality constant as negative i.e. $-k$ . So, the equation will be written as $\dfrac{dt}{dT}=-k\left( T-S \right)$ .
On rearranging the terms, we get $\dfrac{dt}{\left( T-S \right)}=-kdt$
Now, we will integrate on both sides of equation we will get
$\int{\dfrac{dt}{\left( T-S \right)}}=\int{-kdt}$
We know that $\int{\dfrac{1}{a}=\log a}$ so using, this we will get
$\log \left( T-S \right)=-kt+c$ ………………………(1)
Where c is a constant term.
So, here taking T as $100{}^\circ F$ at $t=1$ we get
$\log \left( 100-S \right)=-k+c=c$ ………………..(2) (considering k as constant so, writing it as one term c)
So, putting equation (2) in equation (1), we get
$\log \left( T-S \right)=-kt+\log \left( 100-S \right)$ ……………………….(3)
Now taking $t=5$ and T as $50{}^\circ F$ and putting in equation (3), we get
$\log \left( 50-S \right)=-5k+\log \left( 100-S \right)$ ………………………(4)
Similarly, taking $t=10$ and T as $40{}^\circ F$ and putting in equation (3), we get
$\log \left( 40-S \right)=-10k+\log \left( 100-S \right)$ …………………(5)
From equation (4), we will make constant term k as subject and on rearranging the terms, we get
$\Rightarrow \dfrac{1}{5}\log \dfrac{\left( 50-S \right)}{\left( 100-S \right)}=-k$ ……………………….(6)
Similarly, we will do in equation (5) making k as subject we get as
$\Rightarrow \dfrac{1}{10}\log \dfrac{\left( 40-S \right)}{\left( 100-S \right)}=-k$ …………………..(7)
Now, equating equation (6) and (7), we get
$\Rightarrow \dfrac{1}{10}\log \dfrac{\left( 40-S \right)}{\left( 100-S \right)}=\dfrac{1}{5}\log \dfrac{\left( 50-S \right)}{\left( 100-S \right)}$
On solving, we get
$\Rightarrow \log \dfrac{\left( 40-S \right)}{\left( 100-S \right)}=2\log \dfrac{\left( 50-S \right)}{\left( 100-S \right)}$
So, removing the log from both the sides we get
\[\Rightarrow \dfrac{\left( 40-S \right)}{\left( 100-S \right)}={{\left( \dfrac{\left( 50-S \right)}{\left( 100-S \right)} \right)}^{2}}\] (using the property $n\log m={{m}^{n}}$ )
On taking LCM on LHS side and simplifying the terms we get
\[\Rightarrow \left( 40-S \right)\left( 100-S \right)={{\left( 50-S \right)}^{2}}\]
$\Rightarrow 4000-140S+{{S}^{2}}=2500-100S+{{S}^{2}}$
$\Rightarrow 4000-2500=140S-100S$
$\Rightarrow 1500=40S$
$\Rightarrow \dfrac{1500}{40}=S=37.5{}^\circ $
Thus, the temperature of the surroundings is $37.5{}^\circ F$ .

Note: Remember that here temperature is decreasing at an interval of 5 minutes. So, negative sign will come in proportionality constant in case if plus sign will be there no change in answer. Also, in the line given as “After another 5 minutes, the temperature becomes $40{}^\circ F$” here do not take it as 5. Here is a consecutive decrease of 5minutes when temperature was 50degree. So, time will be $t=5+5=10$ . Instead of this if t equals 5 is taken then there will be change in the answer. So, don’t make this mistake.