Answer
See below
Work Step by Step
Given: $T(0)=80\\T(3)=100\\T_m=180\\T(t_0)=140$
The temperature of the sandals at time $t$ can be expressed as:
$\frac{dT}{dt}=-k(T-180)$
Integrating both sides: $\int \frac{1}{T-180}dT=-\int kdt\\
\rightarrow \ln|T-180|=-kt+c\\
\rightarrow T(t)=c_1e^{-kt}+180$
The initial value condition $T(0)=80$, then $c_1+180=80\\c_1=-100$
Hence, the solution is $T(t)=-100e^{-kt}+180$
Substitute the given condition $T(3)=100$, then:
$100=-100e^{-3t}+180\\
\rightarrow e^{-3k}=\frac{80}{100}=\frac{4}{5}\\
\rightarrow -3k=\ln\frac{4}{5}\\
\rightarrow k=-\frac{1}{3}\ln \frac{4}{5}$
Substitute back the result, we have:
$140=-100e^{t_0\frac{1}{3}\ln\frac{4}{5}}+180\\
\rightarrow e^{t_0\frac{1}{3}\ln\frac{4}{5}}=\frac{2}{5}\\
\rightarrow t_0\frac{1}{3}\ln\frac{4}{5}= \ln\frac{2}{5}\\
\rightarrow t_0\approx12.319$
