Chapter 6.6, Problem 19E

To determine

To calculate: .The waiting time when you can continue driving.

Answer to Problem 19E

The required time is $50$ minute.

Explanation of Solution

Given Information:

Temperature of car overheats is ${280}^{\circ }\text{F}$ .

Temperature of car driven is ${230}^{\circ }\text{F}$ .

Temperature of outside is ${80}^{\circ }\text{F}$ .

Also, cooling rate of car is $r=0.0058$ .

Formula Used:

Newton’s Law of Cooling:

  $T=\left(T_0-T_R\right)e^{-rt}+T_R$

Where, $T_0$ is initial temperature, $T$ is temperature of the body, $T_R$ is surrounding temperature, $r$ is heat transfer coefficient, and $t$ is time.

Power rule of logarithmic functions.

  ${\log }_a{x}^p=p{\log }_{a}x$

Where $a$ is the base, $x$ is the variables and $p$ is the power of variables.

Calculation:

Use Newton’s Law of Cooling formula.

  $T=\left(T_0-T_R\right)e^{-rt}+T_R$

Substitute for $T_0$ , $230$ for $T$ , $80$ for $T_R$ and $0.0058$ for $0.0058$ .

  $230=\left(280-80\right)e^{-\left(0.0058\right)t}+80$

Simplify the above equation.

  $\begin{array}{c}230=\left(280-80\right)e^{-\left(0.0058\right)t}+80\\ 230=200e^{-\left(0.0058\right)t}+80\\ 230-80=200e^{-\left(0.0058\right)t}\\ 150=200e^{-\left(0.0058\right)t}\end{array}$

Simplify the equation.

  $\begin{array}{c}150=200e^{-\left(0.0058\right)t}\\ \frac{150}{200}=e^{-\left(0.0058\right)t}\\ 0.75=e^{-\left(0.0058\right)t}\end{array}$

Take natural logarithm with base $e$ of the above equation.

  ${\ln }_e\left(0.75\right)={\ln }_e{e}^{-\left(0.0058\right)t}$

Now, use power rule of logarithm ${\log }_a{x}^p=p{\log }_{a}x$ in the above equation.

  ${\ln }_e\left(0.75\right)=\left(-0.0058t\right){\ln }_{e}e$

Substitute $1$ for ${\ln }_{e}e$ in the above equations.

  $\begin{array}{c}{\ln }_e\left(0.75\right)=\left(-0.0058t\right)\left(1\right)\\ {\ln }_e\left(0.75\right)=\left(-0.0058t\right)\end{array}$

Substitute $-0.2877$ for ${\ln }_e\left(0.75\right)$ in the above equation.

  $-0.2877=\left(-0.0058t\right)$

Simplify the above equation.

  $\begin{array}{c}-0.2877=\left(-0.0058t\right)\\ t=\frac{-0.2877}{\left(-0.0058\right)}\\ t=49.603\\ t\approx 50\end{array}$

Hence, the required time is $50$ minutes.