Heating and Cooling 1

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.

Show Solution

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

$$\begin{align*} &\Step{1A}{-\frac{1}{K}\ln\left| M-T \right| =t+c,}\\ &\Step{2A}{\ln\left| M-T \right| =-Kt+c,} \\ &\Step{3A}{M-T =ce^{-Kt},} \\ &\Step{4A}{T(t) =M+ce^{-Kt}.}\\ \end{align*}$$

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 $T_0$. $\blacksquare$