Answer
(a). The initial temperature is $T(0)=210$
(b). The temperature after $T(10)=152.95$
(c). $t=28.43$
Work Step by Step
$T(t)=T_s+D_0e^{-kt}$. Whereas, $T(t)$ is a temperature at a time $t$, $T_s$ is the surrounding temperature, $D_0$ is the Initial temperature difference between the object and it's surrounding, $k$ is a positive constant that depends on the type of object.
In this case, $T(t)=65+145e^{-0.05t}$
(a). The initial temperature is $T(0)=65+145e^0=210$
(b). The temperature after $T(10)=65+145e^{-0.5}=152.95$
(c). $100=65+145e^{-0.05}$,
$35=145e^{-0.05}$,
$0.2414=e^{-0.05}$,
$\ln 0.2414=-0.05t$,
$t=28.43$
