Answer
See below
Work Step by Step
a) Given: $T_m=72$
Apply Newton's Law of Cooling:
$\frac{dT}{dt}=-k(T-T_m)\\
\rightarrow \frac{dT}{dt}=-k(T-72)\\
\rightarrow \frac{1}{T-72}\frac{dT}{dt}=-k$
Integrate: $\int \frac{1}{T-72}dT=\int -k.dt\\
\rightarrow \ln(T-72)=-kt+c_1$
Solve for $T$
$T=72+c_1e^{-kt}\\
\rightarrow 150=72+c_1e^{-k}\\
\rightarrow c_1e^{-k}=78$
Find $k$: $-20=-k(150-72)\\
\rightarrow -78k=-20\\
\rightarrow k=0.256$
Find $c_1$:
$78=c_1e^{(-0.256)(1)}\\
\rightarrow c_1=100.76$
Hence, $T(0)=72+100.76e^{-(0.256)(0)})\\
\rightarrow T(0)=72+100.76\\
\rightarrow T(0)=172.76$
The temperature after 10 minutes is:
$T(0)=72+100.76e^{-(0.256)(10)})\\
\rightarrow T(0)=72+7.79\\
\rightarrow T(0)=79.79$
The rate of change after 10 minutes:
$\frac{dT}{dt}=-0.256(79.79-72)\\
\frac{dT}{dt} \approx-2$
