Chapter 7.7, Problem 56PPS

(a)

To determine

To find: The value of $k$ .

Answer to Problem 56PPS

The value of $k=0.4520$ .

Explanation of Solution

Given information:

The given function is $T\left(t\right)=T_s+\left(T_0-T_s\right)e^{-kt}$ .

Calculation:

Calculate the solution

  $\begin{array}{l}{T}_0=180\\ T_s=70\end{array}$

After $10$ minutes the coffee cools to ${140}^0$

  $\begin{array}{l}T\left(t=10\right)={140}^0\\ 70+\left(180-70\right)e^{-10k}=140\\ 110e^{-10k}=70\\ e^{-10k}=\frac{7}{11}\end{array}$

Natural logarithm both sides

  $\begin{array}{c}\ln e^{-10k}=\ln \frac{7}{11}\\ -10k=\ln \frac{7}{11}\\ k=-\frac{1}{10}\ln \frac{7}{11}\\ =0.4520\end{array}$

  

Therefore the required value of $k=0.4520$ .

(b)

To determine

To find: The temperature of the coffee after $20$ minutes.

Answer to Problem 56PPS

The temperature of the coffee after $20$ minute ${70.01}^0$ .

Explanation of Solution

Given information:

The given function is $T\left(t\right)=T_s+\left(T_0-T_s\right)e^{-kt}$ .

Calculation:

Calculate solution.

  $20$

  

Therefore the temperature of the coffee after $20$ minute ${70.01}^0$ .

(c)

To determine

To find: The time temperature of the coffee reach ${75}^0$ .

Answer to Problem 56PPS

The time temperature of the coffee reach ${75}^0$ is $6.8$ min.

Explanation of Solution

Given information:

The given function is $T\left(t\right)=T_s+\left(T_0-T_s\right)e^{-kt}$ .

Calculation:

Calculate the equation.

  $\begin{array}{l}70+\left(180-70\right)e^{-0.4520t}=75\\ 70+110e^{-0.4520t}=75\\ 110e^{-0.4520t}=5\\ e^{-0.4520}=\frac{5}{110}\end{array}$

The natural logarithm of both sides

  $\begin{array}{l}\ln e^{-0.4520t}=\ln \frac{5}{110}\\ -0.4520t=\ln \frac{5}{110}\\ t=-\frac{1}{0.4520}\ln \frac{5}{110}\\ =6.8\end{array}$

Use $TI83$ to calculate to computer value

  

Therefore, the time temperature of the coffee reach ${75}^0$ is $6.8$ min.