Question

A cup of coffee cools according to Newton’s law of cooling (3). Use data from the graph of the temperature $\mathit{T}\left(t\right)$ in figure 1.3.10 to estimate the constants T_m,T_0, and k in a model of the form of a first-order initial- value problem : localid="1663843681637" $\mathit{d}\mathit{T}\mathbf{/}\mathit{d}\mathit{t}\mathbf{=}\mathit{k}\mathbf{(}\mathit{T}\mathbf{-}{\mathit{T}}_{\mathbf{m}}\mathbf{)}$, $\mathit{T}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{T}}_{\mathbf{0}}$.

Short answer

The constants by using the data given in graph are ;

$$\begin{gathered} T_0=180 \\ {T}_m=75 \\ k=-0.256 \end{gathered}$$

Step-by-step solution

Definition of a differential equation, initial -value problem

A differential equation contains derivatives, partial or ordinary derivatives .in applications, it functions as the rate of change, physical quantity.

$dy/dx=f\left(x\right)$

The interval I, containing $x_0$the problem solving an nth order differential equation subjected to n side conditions specified at $x_0$.

Solve:$\frac{d^{n}y}{d{x}^n}=F\left(x,y,y\text{'},.....,y^{(n-1)}\right)$

Subject to:$y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,...,y^{(n-1)}\left(x_0\right)=y_{n-1}$

Where, $y_0,y_1,.....,y_{n-1}$are arbitrary constants called an nth-order initial value problem (IVP).

The values of y(x) and its first (n-1) derivatives at $y_0,y\text{'}\left(x_0\right)=y_1,...,y^{(n-1)}\left(x_0\right)=y_{n-1}$are called initial conditions(IC).

Formula for estimation of temperatures are:

$$\begin{gathered} dy/dx=f\left(x\right) \\ dT/dt=k(T_o-T_m) \end{gathered}$$

Where, ${\text{T}}_{\text{0}}$is the initial temperature (temperature given in graph) and $T_m$ time (in minutes)

Estimation of temperature

$T_0$is temperature, when $t=0$.

From the graph we can estimate that $T_0=180$.

$T_m$is ambient temperature

If we let the coffee to cool down for a long time then the temperature of the coffee will be the ambient temperature

From graph we estimate as 75

$T_m=75$

Now we have to find the temperature of coffee at when it doesn’t change anymore.

Solving using the formula

Now we have to find k using the point and we take arbitrary point from graph as (12.5,100).

Now we can take the slope as follows:

$$\begin{gathered} {\left.\frac{dT}{dt}\right|}_{ t=12.5}=\frac{100-180}{12.5-0} \\ =-6.4 \end{gathered}$$

We substitute in the differential equation to get k,

$$\begin{gathered} {\frac{dT}{dt}}_{ t=12.5}=k\left(T { }_{t=12.5}-T_m\right) \\ -6.4=k\left(100-75\right) \\ k=-0.256 \end{gathered}$$

The temperature and constant are $T_0=180,T_m=75,k=-0.256$.