Question

A hot body, obeying Newton’s law of cooling is cooling down from its peak value ${80}^{\circ }C$to an ambient temperature of ${30}^{\circ }C$. It takes 5 minutes to cool down from ${80}^{\circ }C$to ${40}^{\circ }C$. How much time will it take to cool down from ${62}^{\circ }C$to ${32}^{\circ }C$? (Given $\ln 2=0.693,\ln 5=1.609$)
A. $8.6$ minutes
B. $6.5$ minutes
C. $9.6$ minutes
D. $3.75$ minutes

Answer 1

Hint:We have been given two cases of different temperature ranges in which the temperature of the body is changing. In one of them the time span is also mentioned but what we must see is that the value of the constant used in the equation of Newton’s law of cooling is missing. Hence, we shall first find the value of this constant and then proceed further to calculate the time taken in the second temperature range.

Complete step-by-step solution:
According to Newton’s law of cooling;
$T\left(t\right)=T_f+(T_i-T_f)e^{-kt}$ ……………………. Equation (1)
Where,
$T\left(t\right)=$ temperature of body at time $t$
$T_f=$ final temperature of the body
$T_i=$ initial temperature of the body
$t=$ time taken
$k=$ positive constant of Newton’s law of cooling

In the first case,
We have $T\left(t\right)={40}^{\circ },T_f={30}^{\circ },T_i={80}^{\circ },t=5$minutes.

Substituting these values in equation (1), we get
$\begin{array}{rl}& \Rightarrow {40}^{\circ }={30}^{\circ }+({80}^{\circ }-{30}^{\circ })e^{-k5}\\ & \Rightarrow {10}^{\circ }=\left({50}^{\circ }\right)e^{-5k}\end{array}$

Taking log on both sides,
$\Rightarrow \ln 10=\ln \left(50\right)e^{-5k}$

Now using the logarithmic property, $\ln ab=\ln a+\ln b$ and $\ln a^b=b\ln a$, we get
$\begin{array}{rl}& \Rightarrow \ln 10=\ln \left(50\right)+(-5k)\ln e\\ & \Rightarrow \ln 10=\ln 5+\ln 10-5k\end{array}$
$\{\because \ln e=1\}$

Cancelling $\ln 10$from both sides,
$\Rightarrow \ln 5=5k$
Also, given that $\ln 5=1.609$,
$\begin{array}{rl}& \Rightarrow 5k=1.609\\ & \Rightarrow k={\displaystyle \frac{1.609}{5}}\\ & \Rightarrow k=0.3218\end{array}$

Now, in the second case,
We have $$T\left(t\right)={32}^{\circ },T_f={30}^{\circ },T_i={62}^{\circ },k=0.3218$$

Substituting these values in equation (1) to find time required for cooling, we get
$\begin{array}{rl}& \Rightarrow {32}^{\circ }={30}^{\circ }+({62}^{\circ }-{30}^{\circ })e^{-0.3218t}\\ & \Rightarrow 2=\left(32\right)e^{-0.3218t}\\ & \Rightarrow 2=\left(2^5\right)e^{-0.3218t}\end{array}$
Taking log on both sides,
$\Rightarrow \ln 2=\ln \left(2^5\right)e^{-0.3218t}$

Again, using the logarithmic property, $\ln ab=\ln a+\ln b$ and $\ln a^b=b\ln a$, we get
$\begin{array}{rl}& \Rightarrow \ln 2=\ln 2^5+\ln e^{-0.3218t}\\ & \Rightarrow \ln 2=5\ln 2+(-0.3218t)\end{array}$
$\{\because \ln e=1\}$
$\Rightarrow 0.3218t=4\ln 2$

Given that $\ln 2=0.693$, substituting this value,
$\begin{array}{rl}& \Rightarrow 0.3218t=4\left(0.693\right)\\ & \Rightarrow t={\displaystyle \frac{4\left(0.693\right)}{0.3218}}\\ & \Rightarrow t={\displaystyle \frac{2.772}{0.3218}}\\ & \Rightarrow t=8.61\end{array}$

Therefore, the time taken by body to cool down from ${62}^{\circ }C$ to ${32}^{\circ }C$ is $8.61$ minutes.

Therefore, the correct option is (A) $8.6$ minutes.

Note:
Generally, when a body which is hotter or cooler than the ambient room temperature goes under a change in temperature when placed in a different surrounding. This is due to Newton’s law of cooling which conventionally states that the rate of change of temperature should be proportional to the difference between the temperature of the object and the ambient temperature.