Question

The temperature $T$ of a cooling object drops at a rate proportional to the difference $(T-S)$, where $S$ is constant temperature of surrounding medium. If initially $T={150}^{\circ }C$, find the temperature of the cooling object at any time ${}^{\prime }{t}^{\prime }$.

Answer 1

Let $T$ be the temperature of the cooling object at any time $t$.
Given, $\frac{dT}{dt}\propto (T-S)\Rightarrow \frac{dT}{dt}=-K(T-S)$ where $k$ is negative
$\Rightarrow \frac{dT}{T-S}=-k.dt$
$\Rightarrow \log (T-S)=-kt+\log c$
$\Rightarrow \log (T-S)=-\log c=kt$
$\Rightarrow \log \frac{T-S}{c}=-kt\Rightarrow \log \frac{T-S}{c}=e^{-kt}$
$\Rightarrow T-S=c.e^{-kt}\Rightarrow T=S+c{e}^{-kt}$
When $t=0$ and $T=150$,
$\Rightarrow 150=S+c\Rightarrow c=150-S$
$\therefore$ The temperature of the cooling object at any time ${}^{\prime }{t}^{\prime }$ is $T$.
$=S+(150-S)e^{-kt}$