Chapter 7, Problem 14RE

a.

To determine

To calculate : the temperature of the chocolate after another half hour

Answer to Problem 14RE

The temperature of the chocolate after another half hour is $46.67\text{ degree}$

Explanation of Solution

Given information : The question is

A cup of hot chocolate has temperature ${80}^{\circ }C$ in a room kept at a ${20}^{\circ }C$ . After half an hour the hot chocolate cools to ${60}^{\circ }C$

Calculation :

By Newton’s law of cooling

  $\begin{array}{l}\frac{dT}{dt}=k\left(T-20\right)\\ y=T-20\\ y\left(0\right)=T-20\Rightarrow 80-20=60\\ y\left(t\right)=y\left(0\right)\times e^{kt}\\ y\left(t\right)=60e^{kt}\\ T\left(30\right)=60\\ y\left(30\right)=60-20=40\\ 40=60e^{k\left(30\right)}\\ \frac{2}{3}=e^{k\left(30\right)}\\ \ln \left(\frac{2}{3}\right)=30k\\ k\approx -0.0135\\ y\left(t\right)=60e^{-0.0135\times 60}\\ T\left(t\right)=20+60e^{-0.0135\times 60}\\ T\left(60\right)=46.67\text{ degree}\end{array}$

b.

To determine

When will the chocolate have cooled to ${40}^{\circ }C$

Answer to Problem 14RE

It takes $81.285\min$ to cooled to ${40}^{\circ }C$

Explanation of Solution

Given information : The question is

A cup of hot chocolate has temperature ${80}^{\circ }C$ in a room kept at a ${20}^{\circ }C$ . After half an hour the hot chocolate cools to ${60}^{\circ }C$

Calculation :

From part (a)

  $\begin{array}{l}T\left(t\right)=20+60e^{-0.0135\times t}\\ \text{40=}20+60e^{-0.0135\times t}\\ \frac{1}{3}=e^{-0.0135\times t}\\ \ln \left(\frac{1}{3}\right)=-0.0135t\\ t=81.285\min \end{array}$