Question

question 22
newton’s law of cooling can be represented by the equation
$t(t)=t_0e^{-kt}$
where $t(t)$ is the final temperature in degrees celsius, $t_0$ is the initial temperature in degrees celsius, $t$ is the elapsed time in minutes, and both $e$ & $k$ are constants.
$e = 2.718$
$k = 0.043$
the length of time, in minutes, required for a cup of coffee to cool from $82 \\ ^{circ}$c to $65^{circ}$c is ______.
question 23

Explanation:

Step1: Identify known values

We know \( T(t) = 65^\circ\text{C} \), \( T_0 = 82^\circ\text{C} \), \( k = 0.043 \), and the formula \( T(t)=T_0e^{-kt} \). Substitute the known values into the formula:
\( 65 = 82e^{-0.043t} \)

Step2: Solve for \( t \)

First, divide both sides by 82:
\( \frac{65}{82}=e^{-0.043t} \)
Calculate \( \frac{65}{82}\approx0.7927 \), so:
\( 0.7927 = e^{-0.043t} \)
Take the natural logarithm of both sides:
\( \ln(0.7927)=\ln(e^{-0.043t}) \)
Using the property \( \ln(e^x)=x \), we get:
\( \ln(0.7927)= - 0.043t \)
Calculate \( \ln(0.7927)\approx - 0.232 \), then:
\( - 0.232=-0.043t \)
Divide both sides by \( - 0.043 \):
\( t=\frac{-0.232}{-0.043}\approx5.395 \)

Answer:

Approximately \( 5.4 \) minutes (or more precisely \( \approx5.4 \) depending on rounding, the exact calculation gives around \( 5.4 \))