Chapter 12, Problem 8P

(a)

To determine

Tofind: The recursive formula theta model the soup temperature $T_n$ at the nth minute.

Answer to Problem 8P

  $T_n=T_{n+1}-3e^{0.03n}$

Explanation of Solution

Given:A tureen of soup at a temperature of ${170}^{\circ }F$ is placed on a table in a dining room in which the thermostat is set at ${70}^{\circ }F$

. The soup cools according to the following rule, a special case of Newton’s law of cooling: Each minute, the temperature of the soup declines by 3% of the difference between the soup temperature and the room temperature.

Newton’s Law of Cooling

  $\begin{array}{l}T\left(t\right)=T_s+\left(T_o-T_s\right)e^{kt}\\ T\left(t\right)=\text{temperature at time t}\\ T_s=\text{temperature of surrounding}\\ T_0=\text{initial temperature}\\ k=\text{contant}\end{array}$

According to question,

  $\begin{array}{l}{T}_s={70}^{\circ }\text{F }\left(\text{temperature of surrounding}\right)\\ T_0={170}^{\circ }\text{F }\left(\text{initial temperature}\right)\\ k=0.03\left(\text{contant}\right)\end{array}$

Substitute the value into formula

  $\begin{array}{c}T\left(t\right)=70+\left(170-70\right)e^{0.03t}\\ T_n=70+100e^{0.03n}\\ T_{n+1}=70+100e^{0.03\left(n+1\right)}\\ T_n-T_{n+1}=100e^{0.03n}\left(1-e^{0.03}\right)\\ T_n-T_{n+1}=-3e^{0.03n}\\ T_n=T_{n+1}-3e^{0.03n}\end{array}$

Hence, the recursive formula is $T_n=T_{n+1}-3e^{0.03n}$ .

(b)

To determine

To find: The recursive formula theta model the soup temperature $T_n$ at the nth minute.

Answer to Problem 8P

  $\begin{array}{cc}n& T_n\\ 0& 170\\ 10& 165.9\\ 20& 160.5\\ 30& 153.1\\ 40& 143.1\\ 50& 129.6\\ 60& 111.5\end{array}$

Explanation of Solution

Given: A tureen of soup at a temperature of ${170}^{\circ }F$ is placed on a table in a dining room in which the thermostat is set at ${70}^{\circ }F$

. The soup cools according to the following rule, a special case of Newton’s law of cooling: Each minute, the temperature of the soup declines by 3% of the difference between the soup temperature and the room temperature.

Newton’s Law of Cooling

  $\begin{array}{l}T\left(t\right)=T_s+\left(T_o-T_s\right)e^{kt}\\ T\left(t\right)=\text{temperature at time t}\\ T_s=\text{temperature of surrounding}\\ T_0=\text{initial temperature}\\ k=\text{contant}\end{array}$

The recursive formula is $T_n=T_{n+1}-3e^{0.03n}$

Using recursive formula to make table for 10 minutes increment from $n=0\text{ to }n=60$

  $\begin{array}{cc}n& T_n\\ 0& 170\\ 10& 165.9\\ 20& 160.5\\ 30& 153.1\\ 40& 143.1\\ 50& 129.6\\ 60& 111.5\end{array}$

(c)

To determine

To graph: The recursive model the soup temperature $T_n$ at the nth minute.

Explanation of Solution

Given: A tureen of soup at a temperature of ${170}^{\circ }F$ is placed on a table in a dining room in which the thermostat is set at ${70}^{\circ }F$

. The soup cools according to the following rule, a special case of Newton’s law of cooling: Each minute, the temperature of the soup declines by 3% of the difference between the soup temperature and the room temperature.

Newton’s Law of Cooling

  $\begin{array}{l}T\left(t\right)=T_s+\left(T_o-T_s\right)e^{kt}\\ T\left(t\right)=\text{temperature at time t}\\ T_s=\text{temperature of surrounding}\\ T_0=\text{initial temperature}\\ k=\text{contant}\end{array}$

The recursive formula is $T_n=T_{n+1}-3e^{0.03n}$

Using recursive formula to make table for 10 minutes increment from $n=0\text{ to }n=60$

  $\begin{array}{cc}n& T_n\\ 0& 170\\ 10& 165.9\\ 20& 160.5\\ 30& 153.1\\ 40& 143.1\\ 50& 129.6\\ 60& 111.5\end{array}$

Using above table and graphing utility to draw the graph.

  

After long term, graph is approaching the line T=70.

The temperature of soup is equal to surrounding temperature.