Chapter 7.4, Problem 33E

a.

To determine

To find: An exponential regression equation.

Answer to Problem 33E

An exponential regression equation for the $\left(t,T-T_s\right)$

  $T-T_s=79.47\times {\left(0.932\right)}^t$ .

Explanation of Solution

Given information: A temperature probe is removed from a cup of coffee and placed in water whose temperature is ${10}^{\circ }C$ . The data in given table were collected over the next $30$ seconds.

Experimental Data
Time(sec)  Temperature $\left(C^{\circ }\right)$ $T-T_s\left({}^{\circ }C\right)$
$2$ $80.47$    $70.47$
$5$ $69.39$ $59.39$
$10$ $49.66$ $39.66$
$15$ $35.26$ $25.26$
$20$ $28.15$ $18.15$
$25$ $23.56$ $13.56$
$30$ $20.62$ $10.62$

Formula used: Exponential Regression equation

  $y=a\times b^x$ .

Calculation:

Exponential Regression equation

  $y=a\times b^x$ .

An exponential regression equation for the $\left(t,T-T_s\right)$

  $T-T_s=79.47\times {\left(0.932\right)}^t$

b.

To determine

To find: A model for $\left(t,T\right)$ data.

Answer to Problem 33E

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 coffee and placed in water whose temperature is ${10}^{\circ }C$ . The data in given table were collected over the next $30$ seconds.

Experimental Data
Time(sec)  Temperature $\left(C^{\circ }\right)$ $T-T_s\left({}^{\circ }C\right)$
$2$ $80.47$    $70.47$
$5$ $69.39$ $59.39$
$10$ $49.66$ $39.66$
$15$ $35.26$ $25.26$
$20$ $28.15$ $18.15$
$25$ $23.56$ $13.56$
$30$ $20.62$ $10.62$

Formula used: Exponential Regression equation

  $y=a\times b^x$ .

Calculation: An exponential regression equation for the $\left(t,T-T_s\right)$

  $T-T_s=79.47\times {\left(0.932\right)}^t$ .

Given

  $\begin{array}{l}{T}_s=10\\ T-T_s=79.47\times {\left(0.932\right)}^t\\ T=10+79.47\times {\left(0.932\right)}^t\end{array}$

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: the temperature probe will read ${12}^{\circ }C$ .

Answer to Problem 33E

the temperature probe will read ${12}^{\circ }C$ at $52.3$ second.

Explanation of Solution

Given information: A temperature probe is removed from a cup of coffee and placed in water whose temperature is ${10}^{\circ }C$ . The data in given table were collected over the next $30$ seconds.

Experimental Data
Time(sec)  Temperature $\left(C^{\circ }\right)$ $T-T_s\left({}^{\circ }C\right)$
$2$ $80.47$    $70.47$
$5$ $69.39$ $59.39$
$10$ $49.66$ $39.66$
$15$ $35.26$ $25.26$
$20$ $28.15$ $18.15$
$25$ $23.56$ $13.56$
$30$ $20.62$ $10.62$

An exponential regression equation for the $\left(t,T-T_s\right)$

  $T-T_s=79.47\times {\left(0.932\right)}^t$

Given

  $\begin{array}{l}{T}_s=10\\ T-T_s=79.47\times {\left(0.932\right)}^t\\ T=10+79.47\times {\left(0.932\right)}^t\end{array}$

temperature probe will read ${12}^{\circ }C$

  $\begin{array}{l}12=10+79.47\times {\left(0.932\right)}^t\\ 79.47\times {\left(0.932\right)}^t=2\\ {\left(0.932\right)}^t=\frac{2}{79.47}\\ t\ln \left(0.932\right)=\ln \left(\frac{2}{79.47}\right)\\ t=52.3\end{array}$

d.

To determine

To find: The temperature when the temperature probe was removed.

Answer to Problem 33E

The temperature when the temperature probe was removed

  $T={89.47}^{\circ }C$ .

Explanation of Solution

Given information: A temperature probe is removed from a cup of coffee and placed in water whose temperature is ${10}^{\circ }C$ . The data in given table were collected over the next $30$ seconds.

Experimental Data
Time(sec)  Temperature $\left(C^{\circ }\right)$ $T-T_s\left({}^{\circ }C\right)$
$2$ $80.47$    $70.47$
$5$ $69.39$ $59.39$
$10$ $49.66$ $39.66$
$15$ $35.26$ $25.26$
$20$ $28.15$ $18.15$
$25$ $23.56$ $13.56$
$30$ $20.62$ $10.62$

An exponential regression equation for the $\left(t,T-T_s\right)$

  $T-T_s=79.47\times {\left(0.932\right)}^t$ .

Given

  $\begin{array}{l}{T}_s=10\\ T-T_s=79.47\times {\left(0.932\right)}^t\\ T=10+79.47\times {\left(0.932\right)}^t\end{array}$

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

  $\begin{array}{l}T=10+79.47\times {\left(0.932\right)}^t\\ t=0\\ T=10+79.47\times {\left(0.932\right)}^0\\ T={89.47}^{\circ }C\end{array}$