Question

A hot body obeying Newton's law of cooling is cooled down from its peak value $${80}^{\circ }C$$ to an ambient temperature of ${30}^{\circ }C$. It takes $5min.$ in cooling down from ${80}^{\circ }C\text{ to }{40}^{\circ }C$. How much time will it take to cool down from ${62}^{\circ }C\text{ to }{32}^{\circ }C$
(given $\ln 2=0.693,\ln 5=1.609$)
a. $9.6min.$
b. $3.75min.$
c. $8.6min.$
d. $6.5min.$

Answer 1

Hint In this question, the only formula that will be used is Newton's law of cooling which is $({\theta }_t-{\theta }_o)=({\theta }_p-{\theta }_o)e^{-kt}$
Here , ${\theta }_t$ is the temperature at time t,
${\theta }_o$ is the temperature of surroundings,
${\theta }_p$ is the peak temperature and
$k$ is the constant
We will first determine the unknown value of $k$ and then use this law again to find the time according to new conditions.

 Complete step-by-step solution:
Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperature between the body and its surroundings.
Here the temperature of surroundings or we can say, the ambient temperature is ${30}^{\circ }C$. So, ${\theta }_{\circ }={30}^{\circ }C$
The peak temperature from which it starts cooling down is ${80}^{\circ }C$. It is represented by ${\theta }_p$ . So, ${\theta }_p={80}^{\circ }C$ body cools down from peak temperature after a certain time interval. In this case, the time interval is $5min.$ or $50\times 60=300s$ and temperature ${\theta }_t={40}^{\circ }C$
Newton's law of cooling is mathematically expressed as
$({\theta }_t-{\theta }_o)=({\theta }_p-{\theta }_o)e^{-kt}$
Substituting the values, we get
$$\begin{gathered} 40-30=(80-30)e^{-kt} \\ 10=50e^{-300k} \\ {e}^{300k}=5 \end{gathered}$$
Taking $\log$ both sides,we have
$$\begin{gathered} \ln (e^{300k})=\ln 5 \\ 300k=\ln 5 \\ k={\displaystyle \frac{ln5}{300}} \\ k={\displaystyle \frac{0.609}{300}} \end{gathered}$$
Now, we are asked to calculate the time in which the body will cool down. The surrounding temperature and constant k will remain the same.
Let the unknown time be t
We have
$$\begin{gathered} {\theta }_p={62}^{\circ }C \\ {\theta }_t={32}^{\circ }C \\ {\theta }_o={30}^{\circ }C \\ k={\displaystyle \frac{1.609}{300}} \end{gathered}$$
Using Newton's law of cooling, we have
$\Rightarrow {\theta }_t-{\theta }_o=({\theta }_p-{\theta }_o)e^{-kt}$
$\Rightarrow 32-30=(62-30)e^{-kt}$
$\Rightarrow e^{-kt}=16$
Taking $\log$ both sides
$\ln (e^{-kt})=\ln 16$
$\Rightarrow kt=4\ln 2$
$\Rightarrow t=4{\displaystyle \frac{ln2}{k}}$
$\Rightarrow t={\displaystyle \frac{4\times 0.693\times 300}{1.609}}=516.84s$
$\Rightarrow t=8.614min$
So, option (c) is correct .

Note:- You should be very careful with calculations and should be well versed with laws related to logarithm. Moreover, you should precisely know which physical quantities are represented by ${\theta }_p,{\theta }_t,{\theta }_n,t$