Chapter 4.6, Problem 28E

To determine

To evaluate: The temperature of boiled water 10 min after the point of time when the temperature is $75°\text{C}$ and sketch the graph of temperature function.

Answer to Problem 28E

The temperature of boiled water 10 min after the point of time when the temperature is $75°\text{C}$ is $63°\mathrm{C}$.

Explanation of Solution

Given:

The room temperature is $20°\text{C}$.

The temperature of the boiled water reduces from $100°\text{C}$ to $75°\text{C}$ after 15 min.

Formula used: The Newton’s Law of Cooling.

$T\left(t\right)=T_s+D_0{e}^{-kt}$ (1)

Where,

• $T_s$ is the temperature of the surroundings.
• $D_0$ is the initial difference between the temperature of the object and its surroundings.
• $k$ is a positive constant.

Calculation:

The room temperature $T_s$ is $20°\text{C}$, therefore the initial difference in the temperature is given below,

$\begin{array}{c}{D}_o=100°\text{C}-20°\text{C}\\ =80°\text{C}\end{array}$

Substitute $20°\text{C}$ for $T_s$ and $80°\text{C}$ for $D_0$ in equation (1),

$T\left(t\right)=20+80\left(e^{-kt}\right)$ (2)

The temperature of the boiled water reduces from $100°\text{C}$ to $75°\text{C}$ after 15 min.

Substitute 75 for $T\left(t\right)$ and 15 for $t$ in equation (2), to find the constant $k$,

$\begin{array}{c}75=20+80\left(e^{-k\left(15\right)}\right)\\ 55=80\left(e^{-15k}\right)\\ \frac{55}{80}=e^{-15k}\end{array}$

Simplify the above expression by taking $\ln$ for each side,

$\begin{array}{c}\ln \frac{11}{16}=\ln \left(e^{-15k}\right)\\ -0.374693449=-15k\left(\because \ln \left(e^a\right)=a\right)\\ k=\frac{-0.374693449}{-15}\\ k=0.024979563\end{array}$

The temperature of boiled water is $75°\text{C}$ at $t=15$ and to find the temperature of boiled water 10 min after the point of time when the temperature is $75°\text{C}$, the value of $t$ is given below,

$\begin{array}{c}t=15+10\\ =25\end{array}$

Substitute 0.024979563 for $k$ and 25 for $t$ in equation (2), to

$\begin{array}{c}T\left(25\right)=20+80\left(e^{-\left(0.024979563\right)\left(25\right)}\right)\\ =20+80\left(e^{-0.624489}\right)\\ =20+80\left(0.535535\right)\end{array}$

Simplify the above expression,

$\begin{array}{c}T\left(25\right)=20+42.8424\\ =62.8424\\ \approx 63\end{array}$

Thus, the temperature of boiled water 10 min after the point of time when the temperature is $75°\text{C}$ is $63°\mathrm{C}$.

The temperature function is given below,

$T\left(t\right)=20+80\left(e^{-0.024979563t}\right)$.

The different value of $T\left(t\right)$ at the various point of time $t$ is given in below table.

$t$	$-50$	$0$	$50$	$100$	$150$	$200$	$250$
$T\left(t\right)$	298.94	$100$	$42.94$	$26.58$	$21.89$	$20.54$	$20.16$

In the above table the function $T\left(t\right)$ denotes the temperature at time $t$. The graph of $T\left(t\right)=20+80\left(e^{-0.024979563t}\right)$ denotes the $T\left(t\right)$ on $y$-axis and $t$ on $x$-axis.

Figure (1)

Thus, the figure (1) shows the graph of the temperature function $T\left(t\right)=20+80\left(e^{-0.024979563t}\right)$.