Chapter 7.4, Problem 31E

i.

To determine

To determine the time taken by the soup to cool to ${35}^{\circ }$ C.

Answer to Problem 31E

The time taken by a soup to cool to ${35}^{\circ }$ C is $17.5$ min.

Explanation of Solution

Given:

A cup of soup cooled from ${90}^{\circ }$ C to ${60}^{\circ }$ C in $10$ min in a room whose temperature was ${20}^{\circ }$ C.

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:

Here $T_s={20}^{\circ }C$ , $T_0={90}^{\circ }C$

At $t=10\min \text{ , }T={60}^{\circ }C$

According to Newton’s Law of Cooling

  $\begin{array}{l}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ 60-20=\left(90-20\right)e^{-10k}\\ 40=70e^{-10k}\\ 10k=\ln \frac{7}{4}\\ k=0.05596\end{array}$

Now , initial temperature $T_0={60}^{\circ }C$ and $T={35}^{\circ }C$

  $\begin{array}{l}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ 35-20=\left(60-20\right)e^{-0.05596t}\\ 15=40e^{-0.00596t}\\ 0.05596t=\ln \frac{40}{15}\\ t=17.5\min \end{array}$

Therefore, the time taken by a soup to cool to ${35}^{\circ }$ C is $17.5$ min.

ii.

To determine

To determine the time taken by soup to cool from ${90}^{\circ }C$ to ${35}^{\circ }C$ .

Answer to Problem 31E

The time taken by soup to cool from ${90}^{\circ }C$ to ${35}^{\circ }C$ is $13.25$ min.

Explanation of Solution

Given:

The cup of ${90}^{\circ }C$ soup is put into a freezer whose temperature is $-{15}^{\circ }C$ .

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:

Here $T_0={90}^{\circ }C$ , $T_s=-{15}^{\circ }C$ and $T={35}^{\circ }C$

According to Newton’s law of cooling

  $\begin{array}{l}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ 35-\left(-15\right)=\left(90-\left(-15\right)\right)e^{-0.05596t}\\ 50=105e^{-0.05596t}\\ e^{0.05596t}=\ln \left(\frac{21}{10}\right)\\ t=\frac{\ln \left(2.1\right)}{0.05596}=13.25\min \end{array}$

Therefore, the time taken by soup to cool from ${90}^{\circ }C$ to ${35}^{\circ }C$ is $13.25$ min.