Answer
a. $T=65+120e^{-0.1438t}$
b. after about 8 minutes.
Work Step by Step
a. We are given
$T_0=185^{\circ}F, C=65^{\circ}F, T(2)=155^{\circ}F$
We have
$155=65+(185-65)e^{2k}$
which gives
$k=\frac{ln(90/120)}{2}\approx-0.1438$
Thus the model equation is
$T=65+120e^{-0.1438t}$
b. Letting $T=105$, we have
$65+120e^{-0.1438t}=105$
which gives
$t=-\frac{ln(40/120)}{0.1438}\approx8min$
that is, after about 8 minutes.
