Chapter 7.4, Problem 13E

a.

To determine

To calculate : the temperature after $45$ minutes

Answer to Problem 13E

The temperature after $45$ minutes is

Explanation of Solution

Given information : The question is

If the temperature of the turkey is ${150}^{\circ }F$

Calculation : the temperature after $45$ minutes is,

  $\begin{array}{l}\frac{dT}{dt}=k\left(T-T_a\right)\\ {\displaystyle \int \frac{1}{T-T_a}dT={\displaystyle \int kdt}}\\ u=T-T_a\\ du=dT\\ {\displaystyle \int u^{-1}du=\ln \left|u\right|}\\ \ln \left|T-T_a\right|=kt+C\\ \left|T-T_a\right|=e^{kt+C}=C{e}^{kt}\\ when\text{ T}\ge {\text{T}}_a\\ T\left(t\right)=C{e}^{kt}+{\text{T}}_a\\ when\text{ T}\le {\text{T}}_a\\ T-T_a=C{e}^{kt}\\ T\left(t\right)=-C{e}^{kt}+T_a\\ T\left(t\right)=C{e}^{kt}+{\text{T}}_a\left\{\begin{array}{l}{\text{where T}}_a={35}^{\circ }F;t=30\text{minutes ;}T\left(t\right)={150}^{\circ }F\\ T\left(0\right)=185\end{array}\right\}\\ 150=110e^{k30}+75\\ \frac{150-75}{110}=e^{k30}\\ \frac{1}{30}\ln \frac{75}{110}=k\\ k\approx -0.0128\\ T\left(t\right)\approx 110e^{-0.0128}+75\\ T\left(45\right)\approx {136.9292}^{\circ }F\end{array}$

b.

To determine

When will the turkey have cooled to ${100}^{\circ }F$

Answer to Problem 13E

It takes $116.0549\min utes$ cooled to ${100}^{\circ }F$

Explanation of Solution

Given information : The question is

A roast turkey is taken from an oven when its temperature has reached ${185}^{\circ }F$ and its placed on the table in a room where the temperature is ${75}^{\circ }F$

Calculation : from the part (a)

  $\begin{array}{l}T\left(t\right)\approx 110e^{-0.0128}+75\\ 100\approx 110e^{-0.0128}+75\\ \frac{100-75}{110}=e^{-0.0128}\\ \frac{1}{-0.128}\ln \frac{25}{110}=t\\ t\approx 116.0549\min utes\end{array}$