Chapter 7.2, Problem 28E

a.

To determine

To find: The differential equation.

Answer to Problem 28E

  $\frac{dT}{dt}=-\frac{1}{50}\left(T-20\right)$

Explanation of Solution

Given information:

The temperature of the coffee initially was ${95}^{\circ }\text{ C}$ and room temperature is ${20}^{\circ }\text{ C}$ .

The coffee cools at $1^{\circ }\text{ C}$ per minute when its temperature is ${70}^{\circ }\text{ C}$

Formula used :

Newton’s law of cooling: $\frac{dT}{dt}=-K\left(T-T_a\right)$

  $T_a$ is ambient temperature, $K$ is constant and $T\left(t\right)$ is the temperature of the object at time $t$ .

Calculation:

  $\frac{dT}{dt}=-K\left(T-T_a\right)$

  $T_a=20$

So, $\frac{dT}{dt}=-K\left(T-20\right)$

Now, $\frac{dT}{dt}=1$ when $T=70$

i.e.

  $\begin{array}{l}1=-K\left(70-20\right)\\ \Rightarrow K=-\frac{1}{50}\end{array}$

So, $\frac{dT}{dt}=-\frac{1}{50}\left(T-20\right)$

b.

To determine

To sketch: A direction field and use it to draw a solution curve and find the limiting value of temperature.

Explanation of Solution

Given information:

  $\frac{dT}{dt}=-K\left(T-T_a\right)$

At first the temperature of the coffee was ${95}^{\circ }\text{ C}$

Concept Used:

If $\frac{dy}{dx}=F\left(x,y\right)$ , $F\left(x,y\right)$ gives the slope of the solution curve at point $\left(x,y\right)$ . If small line segments with slope $F\left(x,y\right)$ are drawn on several points $\left(x,y\right)$ , the results are called direction field.

Graph:

  

  

The above graph is the solution of the differential equation.

From the graph it appears that $T\left(t\right)=20$ is the limiting value of the temperature.

Interpretation :

The temperature of the coffee at first was ${95}^{\circ }\text{ C}$

Hence, $t=0$ , $T=95$

c.

To determine

To estimate: The temperature of the coffee after $10$ minutes using Euler’s method.

Answer to Problem 28E

The temperature of the coffee after $10$ minutes is ${81.1529524}^{\circ }$ C

Explanation of Solution

Given information:

  $\frac{dT}{dt}=-K\left(T-T_a\right)$

  $h=2$

  $T\left(0\right)=95$

Formula used :

Euler’s Method: Approximate values for the solution of the initial-value problem $y^{\prime }=F\left(x,y\right)$ , $y\left(x_0\right)=y_0$ with the step size $h$ , at $x_n=x_{n-a}+h$ .

  $y_n=y_{n-1}+hF\left(x_{n-1},y_{n-1}\right)n=1,2,\mathrm{3.....}$

Calculation:

  $h=2$

  $T^{\prime }=-\frac{1}{50}\left(T-20\right)$ and $T\left(0\right)=95$

Here, $T^{\prime }=-\frac{1}{50}\left(T-20\right)=F\left(t,T\right)$

  $\begin{array}{l}{T}_0=95\\ T_1=T_0+hF\left(t_0,T_0\right)=95+2\times \left(-1.5\right)=92\\ T_2=T_1+hF\left(t_1,T_1\right)=92+2\times \left(-1.44\right)=89.12\\ T_3=T_2+hF\left(t_2,T_2\right)=89.12+2\times \left(-1.3824\right)=86.3552\\ T_4=T_3+hF\left(t_3,T_3\right)=86.3552+2\times \left(-1.327104\right)=83.700992\\ T_5=T_4+hF\left(t_4,T_4\right)=83.700992+2\times \left(-1.2740198\right)=81.1529524\end{array}$

  $T\left(10\right)=81.1529524$

The temperature of the coffee after $10$ minutes is ${81.1529524}^{\circ }$ C