Problem 1.
If the temperature of the bread is $120^oC$, of the air is $30^oC$ and $K=0.0366$, determine the temperature of the bread $60$ minutes later.
We use the Newton's law of cooling,
\[
T'(t)=K(M-T(t)),
\] where $T(t)$ is the temperature of the bread, $M$ is the temperature of the air and $K\gt 0$ is a constant. Denote $T_0:=T(0)$. The constant function solution of the equation is
\[
T(t)=\underset{\sim}{M}.
\] This is a solution only in case of $T(0)=M$.
Now determine the non-constant solution.
From the differential equation
\[
\frac{1}{K(M-T)}\,dT=1\,dt.
\] Integrating both sides we obtain
\] Using this value of $c$ we obtain
\[
\bbox[lightblue,5px,border:2px solid red]{\color{#800000}{ \bf{T(t)=M+(T_0-M)e^{-Kt}.}}}
\] So
\[
T(60)=30+90e^{-2.1960}=40.0123.
\] It is worth to note that
\[
\lim_{t\to\infty}T(t)=M,
\] independently of the initial value $T_0$. $\blacksquare$
\[
T'(t)=K(M-T(t)),
\] where $T(t)$ is the temperature of the bread, $M$ is the temperature of the air and $K\gt 0$ is a constant. Denote $T_0:=T(0)$. The constant function solution of the equation is
\[
T(t)=\underset{\sim}{M}.
\] This is a solution only in case of $T(0)=M$.
Now determine the non-constant solution.
From the differential equation
\[
\frac{1}{K(M-T)}\,dT=1\,dt.
\] Integrating both sides we obtain
\begin{align*}
&\Step{1A}{-\frac{1}{K}\ln\left| M-T \right| =t+c,}\\
&\Step{2A}{\ln\left| M-T \right| =-Kt+c,} \\
&\Step{3A}{M-T =ce^{-Kt},} \\
&\Step{4A}{T(t) =M+ce^{-Kt}.}\\
\end{align*}
From the initial condition
\[ c=T_0-M.\] Using this value of $c$ we obtain
\[
\bbox[lightblue,5px,border:2px solid red]{\color{#800000}{ \bf{T(t)=M+(T_0-M)e^{-Kt}.}}}
\] So
\[
T(60)=30+90e^{-2.1960}=40.0123.
\] It is worth to note that
\[
\lim_{t\to\infty}T(t)=M,
\] independently of the initial value $T_0$. $\blacksquare$
