Answer
See below
Work Step by Step
According to Newton's Law of Cooling, we have: $T(0)=150\\T(10)=125\\T_m=70\\T(t_0)=100$
The temperature of the plate at time $t$ can be expressed as:
$\frac{dT}{dt}=-k(T-70)\\\rightarrow \frac{dT}{T-70}=-kdt$
Integrate both sides: $\int \frac{dT}{T-70}=\int -kdt\\
\rightarrow \ln|T-70|=-kt +c\\
\rightarrow T(t)=c_1e^{-kt}+70$
Use the initial-value condition $T(0)=150$ to find: $c_1+70=150\\c_1=80$
Thus, the solution is: $T(t)=80e^{-kt}+70$
For $t_0$, we use $T(10)=125$: $80e^{-10k}+70=125\\
\rightarrow e^{-10k}=\frac{55}{80}=\frac{11}{16}\\
\rightarrow -10k=\ln\frac{11}{16}\\
\rightarrow k=-\frac{1}{10}\ln\frac{11}{16}$
Substitute back: $80e^{t_0\frac{1}{10}\ln\frac{11}{16}}+70=100\\
\rightarrow e^{t_0\frac{1}{10}\ln\frac{11}{16}}=\frac{30}{80}=\frac{3}{8}\\
\rightarrow t_0\frac{1}{10}\ln\frac{11}{16}=\ln\frac{3}{8}$
Therefore, the plate reach 100 Celcius degree at: $t_0=10\frac{\ln\frac{3}{8}}{\ln\frac{11}{16}}\approx26.18$
