Chapter 7.4, Problem 34E

a.

To determine

To find: The temperature in this experiment can be modelled as an exponential function of time.

Answer to Problem 34E

this problem deals with cooling, so use Newton’s law of cooling to model the system.

Explanation of Solution

Given information: A temperature probe is removed from a cup of hot chocolate and placed in ice water. The data in given table were collected over the next $30$ seconds.

Experimental Data
Time(sec)  Temperature $\left(C^{\circ }\right)$
$2$    $74.68$
$5$ $61.99$
$10$ $34.89$
$15$ $21.95$
$20$ $15.36$
$25$ $11.89$
$30$ $10.02$

The rate at which an object’s temperature is changing at any given time is roughly proportional to the difference between its temperature and the temperature of the surrounding medium. Since this problem deals with cooling, so use Newton’s law of cooling to model the system.

b.

To determine

To find: The best exponential model.

Answer to Problem 34E

best exponential model

  $T=79.96\times {\left(0.927\right)}^t$ .

Explanation of Solution

Given information: A temperature probe is removed from a cup of hot chocolate and placed in ice water. The data in given table were collected over the next $30$ seconds.

Experimental Data
Time(sec)  Temperature $\left(C^{\circ }\right)$
$2$    $74.68$
$5$ $61.99$
$10$ $34.89$
$15$ $21.95$
$20$ $15.36$
$25$ $11.89$
$30$ $10.02$

Formula used: Exponential Regression equation

  $y=a\times b^x$ .

Calculation: best exponential model

  $\begin{array}{l}T=79.96\times {\left(0.927\right)}^t\\ T=\text{Temperature}\\ \text{t=time}\end{array}$

Graph of the model on a scatter plot of the data

  

$\begin{array}{l}y=\text{Temperature}\\ \text{x=time}\end{array}$

c.

To determine

To find: Time taken the probe will reach ${\text{5}}^{\circ }C$ .

Answer to Problem 34E

the probe will reach ${\text{5}}^{\circ }C$ at $37$ second.

Explanation of Solution

Given information: A temperature probe is removed from a cup of hot chocolate and placed in ice water. The data in given table were collected over the next $30$ seconds.

Experimental Data
Time(sec)  Temperature $\left(C^{\circ }\right)$
$2$    $74.68$
$5$ $61.99$
$10$ $34.89$
$15$ $21.95$
$20$ $15.36$
$25$ $11.89$
$30$ $10.02$

best exponential model

  $\begin{array}{l}T=79.96\times {\left(0.927\right)}^t\\ T=\text{Temperature}\\ \text{t=time}\end{array}$

  $\begin{array}{l}T=79.96\times {\left(0.927\right)}^t\\ 5=79.96\times {\left(0.927\right)}^t\\ {\left(0.927\right)}^t=\ln 0.0625\\ t=\frac{\ln 0.0625}{\ln 0.9273}\\ t=37\end{array}$

d.

To determine

To find: The temperature of the hot chocolate when the probe was removed.

Answer to Problem 34E

The temperature of the hot chocolate when the probe was removed

  $T={79.96}^{\circ }C$ .

Explanation of Solution

Given information: A temperature probe is removed from a cup of hot chocolate and placed in ice water. The data in given table were collected over the next $30$ seconds.

Experimental Data
Time(sec)  Temperature $\left(C^{\circ }\right)$
$2$    $74.68$
$5$ $61.99$
$10$ $34.89$
$15$ $21.95$
$20$ $15.36$
$25$ $11.89$
$30$ $10.02$

best exponential model

  $\begin{array}{l}T=79.96\times {\left(0.927\right)}^t\\ T=\text{Temperature}\\ \text{t=time}\end{array}$

The probe was removed when $\text{time}=0$

  $\begin{array}{l}T=79.96\times {\left(0.927\right)}^t\\ \text{t}=0\\ T=79.96\times {\left(0.927\right)}^0\\ T=79.96\end{array}$