General edit.
If $p(t)$ is the population as a function of the time $t$ and $k$ is the constant of proportionality, the differential equation that models this law is
$$\dfrac{dp}{dt}=kp.\tag{1}$$
To find the solution $p(t)$ integrate $(1)$ and use the given values
$$p(t_0)=p(1970)=3.712\, \mathrm{billion},\tag{2a}$$
$$p(1980)=4.453\, \mathrm{billion},\tag{2b}$$
where $t_0=1970$ is the initial instant. The equation $(1)$ can be integrated by the method of separation of variables, rewriting it as
$$
\begin{equation*}
\frac{dp}{p}=k\, dt\tag{3}
\end{equation*}
$$
and integrating both sides
$$
\begin{eqnarray*}
\int \frac{dp}{p} &=&\int k\, dt \\
\ln p &=&kt+C,\tag{4}
\end{eqnarray*}
$$
where $C$ is the constant of integration. The equation $(4)$ can be rewritten in different forms, such as
$$
\begin{eqnarray*}
p(t)&=&e^{kt+C}=Be^{kt}\tag{5a} \\
p(t) &=&Ae^{k(t-t_0)}=Ae^{k(t-1970)},\tag{5b}
\end{eqnarray*}
$$
where $A$ and $B$ are constants related to $C$. To proceed we could use either $(5\mathrm{a})$ or $(5\mathrm{b})$. Choosing equation $(5\mathrm{b})$, we find $A$ using the initial population $p(t_0)$ $(2\mathrm{a})$:
$$p(1970)=Ae^{k(1970-1970)}=Ae^0=A= 3.712\text{ billion}.\tag{5b'}$$
Now we can find the constant $k$ using the $1980$ population $(2\mathrm{b})$. We get the equation $$4.453=3.712e^{k(1980-1970)},\tag{5b''}$$
whose solution, by applying logarithms, is
$$k=\frac{1}{10}\ln\frac{4.453}{3.712}\approx 1.8201\times 10^{-2}\tag{5b'''}.$$
We thus have the following solution of $(1)$ that satisfies the initial conditions $(2\mathrm{a,b})$:
$$p(t)=3.712e^{1.8201\times 10^{-2}(t-1970)},\tag{6}$$
from which we obtain the Malthus's law prediction for the world population in $2008$:
$$p(2008)=3.712e^{1.8201\times 10^{-2}(2008-1970)}\approx 7.413\, \text{billion}.\tag{7}$$
Since $k>0$ the function $(6)$ expresses an exponential growth.
There are two principles at work here: 1) Bernoulli's Principle, which states that
$$2 g y + v(y)^2 = v(0)^2$$
where $v=dy/dt = \dot{y}$ is the rate at which the fluid is sinking, and $y$ is the height of the fuid. Also, $g$ is the acceleration due to gravity $\approx 9.8 \text{m}/\text{sec}^2$. $v(0)$ is the speed of the fluid exiting the hole at the bottom.
2) $A(y) v(y) = B v(0)$ - this is a statement that the amount of fluid exiting the container is uniform throughout - sort of a conservation principle. This allows us to get a diffrential equation for the height of the fluid at any time:
$$\dot{y}^2 = \frac{2 g y}{\pi^2 y^2/B^2 - 1}$$
Note that I used the fact that the area of the fluid $A(y)$ at height $y$ is $\pi y$ for this container.
In principle, we have a simple ODE which may be expressed in terms of an integral over $y$; the integral, however, is pretty horrid (I get elliptic integrals over imaginary arguments). Nevertheless, we may exploit the fact that the area of the hole at the bottom is very small so that the area of the fluid above the hole at any time is much greater than $B$. We may therefore neglect the $1$ in the denominator and get the approximate DE:
$$\dot{y} = \pm \frac{B}{\pi}\sqrt{\frac{2 g}{y}}$$
We choose the negative sign because $y$ is decreasing. The solution to this equation takes the form
$$y^{3/2} = y(0)^{3/2} - \frac{3 B}{2 \pi} \sqrt{2 g} t$$
You may then find the approximate time at which the container is empty by setting the above to zero.
Best Answer
You need a model for how the temperature decreases. I will guess that the heat loss is proportional to the temperature difference between the coffee and room temperature. Let the coffee temperature be $T$. My model says $\frac {dT}{dt}=k(T-72)$ and $T(0)=210, T(15)=190$. You need to solve the differential equation, use the two pieces of data to evaluate $k$ and the integration constant, then find the time when $T=150$