Question

1. if you pour a cup of coffee that is 200°f, and set it on a desk in a room that is 68°f, and 10 minutes later is it 145°f, what temperature will it be 15 minutes after you originally poured it? (3 points)

Explanation:

Step1: Recall Newton's Law of Cooling

Newton's Law of Cooling is given by $T(t)=T_s+(T_0 - T_s)e^{-kt}$, where $T(t)$ is the temperature of the object at time $t$, $T_s$ is the surrounding - temperature, $T_0$ is the initial temperature of the object, and $k$ is a constant. Here, $T_0 = 200^{\circ}F$, $T_s=68^{\circ}F$. When $t = 10$ minutes, $T(10)=145^{\circ}F$.

Step2: Find the value of $k$

Substitute the values into the formula: $T(10)=68+(200 - 68)e^{-10k}=145$. First, simplify the equation:
$145=68 + 132e^{-10k}$. Then, subtract 68 from both sides: $145−68 = 132e^{-10k}$, so $77 = 132e^{-10k}$. Then, $e^{-10k}=\frac{77}{132}=\frac{7}{12}$. Take the natural logarithm of both sides: $\ln(e^{-10k})=\ln(\frac{7}{12})$. Using the property $\ln(e^x)=x$, we get $- 10k=\ln(\frac{7}{12})$. Solve for $k$: $k=-\frac{1}{10}\ln(\frac{7}{12})=\frac{1}{10}\ln(\frac{12}{7})$.

Step3: Find the temperature at $t = 15$ minutes

Now, use the formula $T(t)=68+(200 - 68)e^{-kt}$ with $t = 15$ and $k=\frac{1}{10}\ln(\frac{12}{7})$.
$T(15)=68+132e^{-15\times\frac{1}{10}\ln(\frac{12}{7})}=68+132e^{-\frac{3}{2}\ln(\frac{12}{7})}$.
Since $a\ln(b)=\ln(b^a)$ and $e^{\ln(x)} = x$, we have $e^{-\frac{3}{2}\ln(\frac{12}{7})}=e^{\ln((\frac{7}{12})^{\frac{3}{2}})}=(\frac{7}{12})^{\frac{3}{2}}$.
$T(15)=68 + 132\times(\frac{7}{12})^{\frac{3}{2}}$.
$(\frac{7}{12})^{\frac{3}{2}}=\sqrt{\frac{7^3}{12^3}}=\sqrt{\frac{343}{1728}}\approx0.444$.
$132\times0.444 = 58.608$.
$T(15)=68+58.608\approx126.6^{\circ}F$.

Answer:

$126.6^{\circ}F$