Chapter 11.6, Problem 56E

a.

To determine

Find the value of $C$ .

Answer to Problem 56E

  $324hours$

Explanation of Solution

Given information:

Temperature According to Newton's Law of Cooling, the temperature $T$ of an object at time $t$ satisfies the equation $\ln \left|T-T_0\right|=-kt+C$ where $T_0$ is the temperature of the surrounding medium and $k$ and Care constants. Suppose a cup of coffee with temperature of $180°F$ is placed in $72°F$ room at time $t=0$ .

Calculation:

Consider the given equation,

  $\ln \left|T-T_0\right|=-kt+C$

  $T$ is the temperature of object, $T_o$ is the temperature of surrounding , $t$ is time and $k$ is constant.

  $\begin{array}{l}c=\left|T-T_0\right|+kt\\ substitutethevalues,\\ c=\left|180-72\right|+k\left(0\right)\\ c=108+0\\ c=108\end{array}$

Hence, the value of $C$ is $108$ .

b.

To determine

Find the value of $k$ .

Answer to Problem 56E

  $15$

Explanation of Solution

Given information:

Temperature According to Newton's Law of Cooling, the temperature $T$ of an object at time $t$ satisfies the equation $\ln \left|T-T_0\right|=-kt+C$ where $T_0$ is the temperature of the surrounding medium and $k$ and Care constants. Suppose a cup of coffee with temperature of $180°F$ is placed in $72°F$ room at time $t=0$ .

After $2$ minutes, the temperature of the coffee has dropped to $150°F$

Calculation:

Consider the given equation,

  $\ln \left|T-T_0\right|=-kt+C$

  $T$ is the temperature of object, $T_o$ is the temperature of surrounding , $t$ is time and $k$ is constant.

  $\begin{array}{l}\left|T-T_0\right|-c=-kt\\ -kt=\left|T-T_0\right|-c\\ k=\frac{c-\left|T-T_0\right|}{t}\\ putT=150°F,T_0=72°F,c=108,t=2\\ k=\frac{108-\left|150-72\right|}{2}\\ k=\frac{108-\left|78\right|}{2}\\ k=\frac{30}{2}\\ k=15\end{array}$

Hence, the value of $k$ is $15$ .

c.

To determine

Find the time for the temperature of the coffee to $1$ drop to $100°F$ .

Answer to Problem 56E

  $5.33\min$

Explanation of Solution

Given information:

Temperature According to Newton's Law of Cooling, the temperature $T$ of an object at time $t$ satisfies the equation $\ln \left|T-T_0\right|=-kt+C$ where $T_0$ is the temperature of the surrounding medium and $k$ and Care constants. Suppose a cup of coffee with temperature of $180°F$ is placed in $72°F$ room at time $t=0$ .

How much longer will it take

Calculation:

Consider the given equation,

  $\ln \left|T-T_0\right|=-kt+C$

  $T$ is the temperature of object, $T_o$ is the temperature of surrounding , $t$ is time and $k$ is constant.

  $\begin{array}{l}\left|T-T_0\right|-c=-kt\\ -kt=\left|T-T_0\right|-c\\ t=\frac{c-\left|T-T_0\right|}{k}\\ putT=100°F,T_0=72°F,c=108,k=15\\ t=\frac{108-\left|100-72\right|}{15}\\ t=\frac{108-\left|28\right|}{15}\\ t=\frac{80}{15}\\ t=5.33\min \end{array}$

Hence, the value of $t$ is $5.33\min$ .