Newton's law of cooling relates the temperature of an object with the time. This law is the following:
[tex]T=(T_0-T_r)e^{rt}+T_r[/tex]Where:
[tex]\begin{gathered} T=\text{ temperature at time "t"} \\ T_0=\text{ initial temperature} \\ T_r=\text{ temperature of the surroundings} \\ r=\text{ rate of }cooling \\ t=\text{ time} \end{gathered}[/tex]Since we are asked to determien time we need to solve for "t". To do that we will subtract the temperature of the surrounding from both sides:
[tex]T-T_r=(T_0-T_r)e^{-rt}[/tex]Now, we divide by the factor multiplying "e":
[tex]\frac{T-T_r}{T_0-T_r}=e^{-rt}[/tex]Now, we use natural logarithm on both sides:
[tex]\ln (\frac{T-T_r}{T_0-T_r})=\ln (e^{-rt})[/tex]Now, we use the following property of logarithms on the right side:
[tex]\ln e^x=x[/tex]This means that we can lower the exponent, like this:
[tex]\ln (\frac{T-T_r}{T_0-T_r})=-rt[/tex]Now, we divide both sides by "-r":
[tex]-\frac{1}{r}\ln (\frac{T-T_r}{T_0-T_r})=t[/tex]Now, we substitute the values:
[tex]-\frac{1}{0.067}\ln (\frac{100F-72F}{180F-72F})=t[/tex]Solving the operations:
[tex]20.15\min =t[/tex]Therefore, the time to wait is 20.15 minutes.
