Question

A metal ball is heated to a temperature of ${100}^{\circ }C$ at time $t=0$, then placed in water which is maintained at ${40}^{\circ }C$. If the temperature of the ball is reduced to ${60}^{\circ }C$ in $4$ minutes, find the time at which the temperature of the ball is ${50}^{\circ }C$.
Given data:
Rate of change of temperature of the body is given as
$\frac{dT}{dt}=-k(T-T_s)$
where, $T$ = Temperature of the body,
$T_s$ = Temperature of the surrounding,
$k$ = proportionality constant,
$t$ = time in minutes

Answer 1

Given, $T$ be the temperature, $T_s$ be temperature of surrounding and $t$ be time in minutes

Given data:
$T_s={40}^{\circ }C$,
Condition 1) At $t=0,T={100}^{\circ }C$
and 2) at $t=4,T={60}^{\circ }C$

To find: $t$ when $T={50}^{\circ }C$

$\frac{dT}{dt}=-k(T-40)$
$\Rightarrow \frac{dT}{(T-40)}=-kdt$
Integrating both sides,
$\log (T-40)=-kt+C...\left(1\right)$

using condition (1), at $t=0,T={100}^{\circ }C$
$C=\log \left(60\right)$
$\therefore$ Eq.(1) becomes,
$\Rightarrow \log (T-40)=-kt+\log \left(60\right)$
$\Rightarrow \log \left(\frac{T-40}{60}\right)=-kt...\left(2\right)$
using condition (2),
$\log \left(\frac{60-40}{60}\right)=-4k$
$\Rightarrow \log \frac{20}{60}=-4k$
$\Rightarrow \log \frac{1}{3}=-4k$
$\Rightarrow k=-\frac{1}{4}\log \frac{1}{3}$
$\Rightarrow k=\frac{1}{4}\log \left(3\right)$
$\therefore$ Eq. (2) becomes,

$\Rightarrow \log \frac{T-40}{60}=-\frac{t}{4}\log 3$
$\Rightarrow \frac{t}{4}\log 3=-\log \frac{T-40}{60}$
$\Rightarrow \frac{t}{4}\log 3=\log \frac{60}{T-40}$

So, $t$ when $T={50}^{\circ }C$ is,

$\frac{t}{4}\log 3=\log \frac{60}{50-40}=\log 6$
$\Rightarrow \frac{t}{4}\log 3=\log 6$
$\Rightarrow t=\frac{4\log 6}{\log 3}\text{min}$