Answer
See below
Work Step by Step
a) Given: $T_m=75$
Apply Newton's Law of Cooling:
$\frac{dT}{dt}=-k(T-T_m)\\
\rightarrow \frac{dT}{dt}=-k(T-75)\\
\rightarrow \frac{1}{T-75}\frac{dT}{dt}=-k$
Integrate: $\int \frac{1}{T-75}dT=\int -k.dt\\
\rightarrow \ln(T-75)=-kt+c_1$
Solve for $T$
For 10 minutes: $T=75+c_1e^{-kt}\\
\rightarrow 415=75+c_1e^{-10k}\\
\rightarrow c_1e^{-10k}=340$
For 20 minutes: $T=34+c_1e^{-kt}\\
\rightarrow 347=75+c_1e^{-20k}\\
\rightarrow c_1e^{-20k}=272$
Find $k$:
$\frac{ c_1e^{-10k}}{ c_1e^{-20k}}=\frac{340}{272}\\
\rightarrow e^{10k}=1.25\\
\rightarrow k=0.022$
Obtain: $c_1e^{-10(0.022)}=340\\
0.8c_1=340\\
c_1=425$
Hence, $T(0)=75+425e^{-(0.022)(0)})\\
\rightarrow T(0)=75+425\\
\rightarrow T(0)=500$
The temperature of the furnace is $500 \circ$C
b) When the coal reaches 100:
$100=75+425e^{-0.022t}\\
\rightarrow e^{-0.022t}=\frac{25}{425}\\
\rightarrow -0.022t=\ln(\frac{25}{425})\\
\rightarrow t\approx128.8$ minutes
