Chapter 7.4, Problem 15E

a.

To determine

To calculate : the temperature of drink after $50$ minutes

Answer to Problem 15E

The temperature of drink after $50$ minutes is ${13.3333}^{\circ }C$

Explanation of Solution

Given information : The question is

When a cold drink is taken from the refrigerator, its temperature is $5^{\circ }C$ . After $25$ minutes in a ${20}^{\circ }C$ room its temperature has increased ${10}^{\circ }C$

Calculation : The temperature of drink after $50$ minutes is,

  $\begin{array}{l}\frac{dT}{dt}=k\left(T-T_a\right)\\ {\displaystyle \int \frac{1}{T-T_a}dT={\displaystyle \int kdt}}\\ u=T-T_a\\ du=dT\\ {\displaystyle \int u^{-1}du=\ln \left|u\right|}\\ \ln \left|T-T_a\right|=kt+C\\ \left|T-T_a\right|=e^{kt+C}=C{e}^{kt}\\ when\text{ T}\ge {\text{T}}_a\\ T\left(t\right)=C{e}^{kt}+{\text{T}}_a\\ when\text{ T}\le {\text{T}}_a\\ T-T_a=C{e}^{kt}\\ T\left(t\right)=-C{e}^{kt}+T_a\\ T\left(t\right)=-C{e}^{kt}+{\text{T}}_a\left\{\begin{array}{l}{\text{where T}}_a={20}^{\circ }C;t=25\text{minutes ;}T\left(t\right)={10}^{\circ }C\\ T\left(0\right)=5^{\circ }C\end{array}\right\}\\ 10=-15e^{k25}+20\\ \frac{10-20}{-15}=e^{k25}\\ \frac{1}{25}\ln \frac{10}{15}=k\\ k\approx -0.0162\\ T\left(t\right)\approx -15e^{-0.0162}+20\\ T\left(50\right)\approx {13.3333}^{\circ }C\end{array}$

b.

To determine

To calculate : when the temperature will be ${15}^{\circ }C$

Answer to Problem 15E

At $68$ minutes the temperature will be ${15}^{\circ }C$

Explanation of Solution

Given information : The question is

When a cold drink is taken from the refrigerator, its temperature is $5^{\circ }C$ . After $25$ minutes in a ${20}^{\circ }C$ room its temperature has increased ${10}^{\circ }C$

Calculation : when the temperature will be ${15}^{\circ }C$

  $\begin{array}{l}T\left(t\right)=20-15e^{\frac{\ln \left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)t}{25}}\\ T\left(t\right)=20-15e^{\frac{\ln \left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)t}{25}}={15}^{\circ }C\\ e^{\frac{\ln \left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)t}{25}}\\ \to \ln \left(\frac{2}{3}\right)\approx -0.4055\\ \to \frac{-0.4055}{25}\approx -0.01622\\ e^{-0.01622}\\ T\left(t\right)=20-15e^{-0.01622}={15}^{\circ }C\\ t\approx 67.732\text{ minutes}\end{array}$