Answer
(a) $210^\circ F$
(b) $153^\circ F$
(c) $28$ minutes
Work Step by Step
(a) Given the law equation $T(t)=65+145e^{-0.05t}$, we can identify the initial temperature of the soup at $t=0$ as $T(0)=65+145=210^\circ F$
(b) With $t=10$, we have $T(10)=65+145e^{-0.05\times10}\approx153^\circ F$
(c) Let $T=100$, we have $100=65+145e^{-0.05t}$ which gives $t=-ln(\frac{100-65}{145})/0.05\approx28$ minutes
