College Algebra, Chapter 5, 5.5, Section 5.5, Problem 26

Determine the temperature after 10 min and illustrate by graphing the temperature function. Suppose that a kettle full of water is brought to a boil in a room with temperature $20^{\circ} C$ . After 15 the temperature of the water has decreased from $100^{\circ} C$ to $75^{\circ} C$.




Recall the formula for radioactive decay

$m(t) = m_0 e^{-rt}$ in which $\displaystyle r = \frac{\ln 2}{h}$

where

$m(t)$ = mass remaining at time $t$

$m_0$ = initial mass

$r$ = rate of decay

$t$ = time

Based from the given, we can see that



$
\begin{equation}
\begin{aligned}

Ts =& 20^{\circ} C
\\
\\
T(15) =& 75^{\circ} C
\\
\\
D_0 =& 100^{\circ}C - 20^\circ C = 80^\circ C

\end{aligned}
\end{equation}
$


So,


$
\begin{equation}
\begin{aligned}

75 =& 20 + 80 e^{-k(15)}
&& \text{Subtract each side by } 20
\\
\\
55 =& 80 e^{-k (15)}
&& \text{Divide each side by } 80
\\
\\
\frac{55}{80} =& e^{-15 k}
&& \text{Take $\ln$ of each side}
\\
\\
\ln \left( \frac{55}{80} \right) =& -15 k
&& \text{Recall } \ln e = 1
\\
\\
k =& - \frac{\displaystyle \ln \left( \frac{55}{80} \right) }{15}
&& \text{Divide each side by } -15
\\
\\
k =& 0.0250
&&

\end{aligned}
\end{equation}
$



Therefore, the temperature function will be

$T(t) = 20 + 80 e^{-0.0250 t}$

The temperature after another 10 min is $t = 25$ min, so


$
\begin{equation}
\begin{aligned}

T(25) =& 20 + 80 e^{-0.025 (25)}
\\
\\
T(25) =& 62.84^{\circ} C

\end{aligned}
\end{equation}
$


Hence, the graph is