Answer
(a) $u = T+(u_{0}-T)e^{-kt}$.
(b) $k\tau = \ln2$.
Work Step by Step
1.2.15
Newton's Law of Cooling: $du/dt = -k(u-T)$
Initial Condition: $u(0) = u_{0}$.
(a) Find the temperature, $u$, of the object at any time, $t$, Where $T$ is the ambient temperature.
$\textit{Solution}$(a):
We must find the solution to the differential equation,
(1) $du/dt = -k(u-T)$,
having initial condition,
(2) $u(0) = u_{0}$.
Let us transform (1):
$\frac{du/dt}{u-T} = -k$.
Integrating,
$\ln(u-T) = -kt +C$
$u-T = e^{-kt+C}$
$= e^{C}e^{-kt}$
$= ce^{-kt}$. (3)
At initial condition, $t =0$,
$u-T = c$.
So.
(4) $u-T = (u_{0}-T)e^{-kt}$,
or
(5) $u = T+(u_{0}-T)e^{-kt}$.
(b) Let $\tau$ be the time at which the initial temperature has been reduced by one half. Find the relation between $k$ and $\tau$.
$\textit{Solution}$(b):
From (4),
$\frac{1}{2}(u_{0}-T) = (u_{0}-T)e^{-k\tau}$.
$\frac{1}{2} = e^{-k\tau}$.,
$2 = e^{k\tau}$,
$k\tau = \ln2$.
