Answer
See below
Work Step by Step
Given: $T_m=80e^{-\frac{t}{20}}\\k=\frac{1}{40}$
Apply Newton's cooling law, we have:
$\frac{dT}{dt}=-k(T-T_m)\\
\frac{dT}{dt}=-k(T-80e^{-\frac{t}{20}})\\
\frac{dT}{dt}+kT=80e^{-\frac{t}{20}}$
Since this is a linear equation whose solution can be given by:
$T=e^{-\int kdt}(c_1+\int 80ke^{-\frac{t}{20}}(e^{\int kdt})dt)\\
T=c_1e^{-kt}+\frac{1600ke^{-\frac{t}{20}}}{20k-1}\\
T=c_1e^{-\frac{1}{40}t}-80e^{-\frac{1}{20}t}$
We are given $T(0)=0\circ F$, then
$T(0)=c_1-80=0\\
\rightarrow c_1=80$
Substitute: $T(t)=80e^{-\frac{t}{40}}-80e^{-\frac{t}{20}}$
For $t \rightarrow \infty$, we have: $\lim T(t)=\lim (80e^{-\frac{t}{40}}-80e^{-\frac{t}{20}})=0$
Let $T'(t)=0\\
\rightarrow 4e^{-\frac{t}{20}}-2e^{-\frac{t}{40}}=0\\
\rightarrow e^{\frac{t}{20}-\frac{t}{40}}=2\\
\rightarrow t=50\ln 2\\
\rightarrow t\approx 27.73$
Hence, the maximum temperature is:
$T(t_{max})=80e^{-\frac{40\ln 2}{40}}-80e^{-\frac{40\ln 2}{20}}=20$
