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 we have \( 0.7927 = e^{-0.043t} \)

Take the natural logarithm of both sides:
\( \ln(0.7927)=\ln(e^{-0.043t}) \)
Using the property \( \ln(e^x)=x \), the right - hand side simplifies to \( - 0.043t \). And \( \ln(0.7927)\approx - 0.232 \)

So, \( - 0.232=-0.043t \)

Now, solve for \( t \) by dividing both sides by \( - 0.043 \):
\( t=\frac{- 0.232}{-0.043}\approx5.4 \)

Answer:

\( 5.4 \) (approximate value, depending on the level of precision in calculation steps)