[tex]ln(T- \frac{20}{200})=-0.11t \\ e^{ln(T- \frac{20}{200})}=e^{-0.11t} \\ T- \frac{20}{200} =e^{-0.11t} \\ T-0.1=e^{-0.11t} \\ T=e^{-0.11t}+0.1[/tex]
Then, now that we have solved for T, we can evaluate and solve for t=20 minutes.
[tex]T=e^{-0.11t}+0.1 \\ T=e^{-0.11*20}+0.1 \\ T=e^{-2.2}+0.1 \\ T=0.11+0.1 \\ T=0.21[/tex]