show that $kT_{\text{halftime}} = \ln 2$ with a differential equation the following is the Newton’s Law of Cooling, 
$$\frac{dT}{dt} = - k (T - T_a) $$
where $T(t)$ is the temperature of object at time $t$, $T_a$ is the ambient temperature at $23 ^\circ \text{C}$ and $k$ is a positive constant.
I took values from temp of $800 \text{ ml}$ of water cooling at ambient temperature of $23^\circ \text{C}$ from here
I integrate it to get $T(t) = 23 + Ce^{-kt} $ 
To find arbitrary constant $C$, I substitute $t=0$, $T= 100$, and therefore I got $C= 77$
With that, I can find values of $k$. Since $k$ values are different at every time, I find $k$ values at $t=10,20,30$ and took the average which gives me $$k = 0.033474$$
The particular solution now is - $$T(t) = 23 + 77e^{-0.03347t} $$
I am trying to prove that $$ kT_{\text{halftime}} = \ln 2$$ where $T_{halftime}$ is the time taken for an object to cool to half of its initial temperature difference. 
Initial temperature difference - $100-23 = 77$
Therefore Temperature when time is half of its initial temperature difference is $= 38.5$
Finding $t$ at $T=38.5$
$$15.5 = 77 e^{-0.033474t} $$ 
Therefore, $T_{\text{halftime}} = 47.886$ 
$$kT_{\text{halftime}} = \ln 2$$
$$0.033474 \cdot 47.886 = 1.602 \ne \ln 2$$
May I know where I have went wrong ? Thanks ..
 A: I'm not sure what you mean by the $k$-values being different at every time. As mentioned in your link, $k$ is a constant. Moreover, it is a very precise constant, so approximating it is going to get in the way of what we need to do, in this case.
Also, you don't want the temperature to be half of the initial temperature difference. Rather, you want the temperature to have cooled by that amount, meaning the temperature in question will be $100-\frac{77}2.$ Consequently, your approximation of $T_{halftime}$ is not even close. Again, though, approximating it will only get in our way.
Fortunately, we aren't asked to demonstrate a value for either $k$ or $T_{halftime}.$ Instead, you should consider the equation $$100-\frac{77}2=23+77e^{-kT_{halftime}},$$ or equivalently, $$\frac{77}2=77e^{-kT_{halftime}}.$$ Can you take it from there?
A: Denoting $t_{1/2}$ as halftime, and $T_{1/2}$ as temperature at halftime, where the temperature difference with ambient is half that at the start,, i.e. $\frac {T_{1/2}-T_a}{T_0-T_a}=\frac 12$,  we have:
$$\begin{align}
\frac {dT}{dt}&=-k(T-T_a)\\
\int_{T_0}^{T_{1//2}} \frac {dT}{T-T_a}&=-k\int_0^{t_{1/2}} dt\\
\bigg[\ln (T-T_a)\bigg]_{T_0}^{T_{1/2}}&=-kt_{1/2}\\
\ln\left(\frac {T_{1/2}-T_a}{T_0-T_a}\right)&=-kt_{1/2}\\
\ln\frac 12&=-kt_{1/2}\\
kt_{1/2}&=\ln 2
\end{align}$$
