we have the formula
[tex]T(t)=T_s+D_0e^{-kt}[/tex]
step 1
Find out the value of k
we have
T=150 degrees
Ts=70
To=200
Do=200-70=130
t=10 min
substitute
[tex]\begin{gathered} 150=70+130e^{-10k} \\ 150-70=130e^{-10k} \\ 80=130e^{-10k} \\ \frac{80}{130}=e^{-10k} \end{gathered}[/tex]
Apply ln on both sides
[tex]\ln (\frac{80}{130})=\ln e^{-10k}[/tex][tex]\begin{gathered} \ln (\frac{80}{130})=-10k\cdot\ln e^{} \\ \ln (\frac{80}{130})=-10k \\ k=0.0486 \end{gathered}[/tex]
substitute the value of k in the given formula
[tex]T(t)=T_s+D_0e^{-0.0486t}[/tex]
Now
For t=20 min
Ts=70
To=200
Do=200-70=130
substitute
[tex]\begin{gathered} T(t)=70+130\cdot e^{-0.0486\cdot10} \\ T(t)=150 \end{gathered}[/tex]
The answer is option A
