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 $5\min .$ 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. $9.6\min .$
b. $3.75\min .$
c. $8.6\min .$
d. $6.5\min .$

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 $5\min .$ 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
$
  40 - 30 = (80 - 30){e^{ - kt}} \\
  10 = 50{e^{ - 300k}} \\
  {e^{300k}} = 5$
Taking $\log$ both sides,we have
$ \ln ({e^{300k}}) = \ln 5 \\
  300k = \ln 5 \\
  k = \dfrac{ln 5}{300} \\
  k = \dfrac{0.609}{300} \\
 $
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
$
  {\theta _p} = {62^ \circ }C \\
  {\theta _t} = {32^ \circ }C \\
  {\theta _o} = {30^ \circ }C \\
  k = \dfrac{1.609}{300} \\
 $
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\dfrac{ln 2}{k }$
$ \Rightarrow t = \dfrac{4 \times 0.693 \times 300}{1.609} = 516.84s$
$ \Rightarrow t = 8.614\min $
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 $\theta _{p},\theta _{t},\theta _{n},t$