Chapter 7.4, Problem 32E

i.

To determine

To determine the temperature above room temperature of an ingot of silver $15$ minutes from now.

Answer to Problem 32E

  $15$ minutes from now , the temperature of silver above room temperature is ${53.455}^{\circ }C$ .

Explanation of Solution

Given:

The temperature of an ingot of silver is ${60}^{\circ }C$ above room temperature right now. Twenty minutes ago , it was ${70}^{\circ }C$ above room temperature.

Concept Used:

Newton’s Law of Cooling If $T$ is the temperature of the object at time $t$ , and $T_s$ is the surrounding temperature, then

  $\frac{dT}{dt}=-k\left(T-T_s\right)$

Its solution , by the law of exponential change ,is

  $T-T_s=\left(T_0-T_s\right)e^{-kt}$ ,

Where $T_0$ is the temperature at time $t=0$ .

Calculation:

Thetemperature of silver above room temperature : $T-T_s=60$

Twenty minutes ago, the temperature of silver above room temperature : $T_0-T_s=70$

Now, according to Newton’s law of cooling

  $\begin{array}{l}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ 60=70e^{-20k}\\ k=0.0077\end{array}$

Now ,considering current temperature of silver above room temperature as a initial temperature that is

  $T_0-T_s=60$ , $15$ minutes from now , the temperature of silver above room temperature is

  $\begin{array}{l}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ T-T_s=60e^{-0.0077\times 15}={53.455}^{\circ }C\end{array}$

Therefore, $15$ minutes from now , the temperature of silver above room temperature is ${53.455}^{\circ }C$

ii.

To determine

To determine the temperature above room temperature of an ingot of silver $2$ hours from now.

Answer to Problem 32E

The temperature above room temperature of an ingot of silver $2$ hours from now is ${23.18}^{\circ }C$ .

Explanation of Solution

Given:

The temperature of an ingot of silver is ${60}^{\circ }C$ above room temperature right now. Twenty minutes ago , it was ${70}^{\circ }C$ above room temperature.

Concept Used:

Newton’s Law of Cooling

If $T$ is the temperature of the object at time $t$ , and $T_s$ is the surrounding temperature, then

  $\frac{dT}{dt}=-k\left(T-T_s\right)$

Its solution , by the law of exponential change ,is

  $T-T_s=\left(T_0-T_s\right)e^{-kt}$ ,

Where $T_0$ is the temperature at time $t=0$ .

Calculation:

Since from part (a) , $k=0.0077$

considering current temperature of silver above room temperature as a initial temperature that is

  $T_0-T_s=60$

Let the temperature of silver above room temperature be $T-T_s$

According to Newton’s law of cooling

  $\begin{array}{l}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ T-T_s=60e^{-0.0077\times 120}={23.18}^{\circ }C\end{array}$

Therefore, the temperature above room temperature of an ingot of silver $2$ hours from now is ${23.18}^{\circ }C$ .

iii.

To determine

To determine the time when the silver be ${10}^{\circ }C$ above room temperature.

Answer to Problem 32E

The time when the silver be ${10}^{\circ }C$ above room temperature is $232.69\min$ .

Explanation of Solution

Given:

The temperature of an ingot of silver is ${60}^{\circ }C$ above room temperature right now. Twenty minutes ago , it was ${70}^{\circ }C$ above room temperature.

Concept Used:

Newton’s Law of Cooling

If $T$ is the temperature of the object at time $t$ , and $T_s$ is the surrounding temperature, then

  $\frac{dT}{dt}=-k\left(T-T_s\right)$

Its solution , by the law of exponential change ,is

  $T-T_s=\left(T_0-T_s\right)e^{-kt}$ ,

Where $T_0$ is the temperature at time $t=0$ .

Calculation:

It is given that the temperature of silver above room temperature is $T-T_s={10}^{\circ }C$

considering current temperature of silver above room temperature as a initial temperature that is $T_0-T_s={60}^{\circ }C$ .

According to Newton’s Law of cooling

  $\begin{array}{l}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ 10=60e^{-0.0077t}\\ 0.0077t=\ln 6\\ t=232.69\min \end{array}$

Therefore, the time when the silver be ${10}^{\circ }C$ above room temperature is $232.69\min$ .