Answer
a. $T=70+380e^{-0.1t}$
b. $121.7^\circ F$
c. $17min$
Work Step by Step
Using Newton's Law of Cooling $T=C+(T_0-C)e^{kt}$, we have:
a. $T_0=450^\circ F, C=70^\circ F, t=5min, T=300^\circ F$, thus $300=70+(450-70)e^{5k}$ and $e^{5k}=\frac{230}{380}\approx0.605$ which gives $k=\frac{ln(0,605)}{5}\approx-0.1$. The model equation is then $T=70+380e^{-0.1t}$
b. With $t=20min$, we have $T=70+380e^{-0.1(20)}\approx121.7^\circ F$
c. Letting $T=140$, we have $70+380e^{-0.1t}=140$ and $e^{-0.1t}=\frac{70}{380}\approx0.184$; thus $t=-\frac{ln(0.184)}{0.1}\approx17min$
