Chapter 8.9, Problem 5ST

a.

To determine

To find: the temperature of the water $0.24$ min after the heating began.

Answer to Problem 5ST

The temperature of water is ${33.22}^{\circ }\text{C}$ .

Explanation of Solution

Given information:

The table provided in the question,

Time in minutes	$0.0$	$0.1$	$0.2$	$0.3$	$0.4$
Temperature	$30.2$	$31.5$	$32.7$	$34.0$	$35.1$

Calculation: let $x$ represents the temperature,

Time	temperature
$0.2$	$32.7$
$0.24$	$x$
$0.3$	$34.0$

Now, the difference,

  $\begin{array}{l}d=\frac{0.04}{.1}\left(1.3\right)\\ d=.52\text{ [simplify]}\end{array}$

Now, to find the value oftemperature,

  $\begin{array}{l}x=32.7+d\\ x=32.7+.52\text{ [substitute]}\\ x={33.22}^{\circ }\text{C [add]}\end{array}$

Thus, Thetemperature is ${33.22}^{\circ }\text{C}$ .

b.

To determine

To find: the time when the temperature rise to ${32.1}^{\circ }C$ .

Answer to Problem 5ST

After $0.15$ min the temperature rise to ${32.1}^{\circ }C$ .

Explanation of Solution

Given information:

The table provided in the question,

Time in minutes	$0.0$	$0.1$	$0.2$	$0.3$	$0.4$
Temperature	$30.2$	$31.5$	$32.7$	$34.0$	$35.1$

Calculation: let $x$ represents the time,

Time	temperature
$0.1$	$31.5$
$x$	$32.1$
$0.2$	$32.7$

Now, the difference,

  $\begin{array}{l}\frac{d}{.1}=\frac{0.6}{1.2}\\ d=0.05\text{ [simplify]}\end{array}$

Now, to find the value of time,

  $\begin{array}{l}x=0.1+d\\ x=0.1+0.05\text{ [substitute]}\\ x=0.15\text{ [add]}\end{array}$

Thus, After $0.15$ min the temperature rise to ${32.1}^{\circ }C$ .