Question

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

Answer 1

Let $T$ be the temperature of the cooling object at any time $t$.
$\frac{dT}{dt}\times (T-S)$

$\Rightarrow \frac{dT}{dt}=K(T-S)$

$\Rightarrow T-S=c{c}^{kt}$,
where $k$ is negative

$\Rightarrow T=S+C{e}^{kt}$
When $t=0,T=150$

$\Rightarrow 150=S+C$

$\Rightarrow C=150-S$
$\therefore$ the temperature of the cooling object at any time is
$T=S+(150-S)e^{kt}$