Question

The ambient temperature in (3) could be a function of time t. Suppose that in an artificially controlled environment, T_m(t)is periodic with a 24-hour period, as illustrated in Figure 1.3.11.

Derive a mathematical model for the temperature T(t)of a body within this environment.

Short answer

The mathematical model is $\frac{dT}{dt}=k\left(T-80+30\cos \left(\frac{\pi }{12}t\right)\right),t>0$.

Step-by-step solution

Newton’s cooling law

Newton's law of cooling warming of an object: $\frac{dT}{dt}=k\left(T-T_m\right)$

Where,

T is the temperature of the body in the environment.

k is the proportional constant.

Tm is the ambient temperature of the surroundings of the body.

Find value of:

Ambient temperature,$T_m=d-A\cos \left(\omega t\right)$

Here d is the average value of T_mfrom the given graph.

$$\begin{gathered} d=\frac{\text{maximum of the given graph}+\text{minimum of the given graph}}{2} \\ =\frac{110+50}{2} \\ =\frac{160}{2} \\ =80 \end{gathered}$$

Find amplitude & frequency:

Let A be the amplitude of given graph.

$A=\frac{\text{maximum of the given graph}-\text{minimum of the graph}}{2}$

$$\begin{gathered} =\frac{110-50}{2} \\ =\frac{60}{2} \\ =30 \end{gathered}$$

And $\omega$is the frequency,

$$\begin{gathered} \omega =\frac{2\pi }{\text{Period}} \\ =\frac{2\pi }{24} \\ =\frac{\pi }{12} \end{gathered}$$

Find model:

Substitute $d=80,A=30$and $\omega =\frac{\pi }{12} in T_m=d-A\cos \left(\omega t\right)$,

$$\begin{gathered} T_{mt}=d-A\cos \left(\omega t\right) \\ =80-30\cos \left(\frac{\pi }{12}t\right) \end{gathered}$$

Substitute $T_m=80-30\cos \left(\frac{\pi }{12}t\right)$in $\frac{dT}{dt}=k\left(T-T_m\right)$,

$$\begin{gathered} \frac{dT}{dt}=k\left(T-T_m\right) \\ =k\left(T-\left(80-30\cos \left(\frac{\pi }{12}t\right)\right)\right) \\ =k\left(T-80+30\cos \left(\frac{\pi }{12}t\right)\right) \end{gathered}$$

Hence, the mathematical model is, $\frac{dT}{dt}=k\left(T-80+30\cos \left(\frac{\pi }{12}t\right)\right),t>0$.