Differential Equation on Temperature Change So I have a question which I have been struggling to solve and from what I can tell, it uses differential equation but I am just unsure of how to implement the question into a differential equation and then solve it. Any help on the matter would be greatly appreciated. Thank you.
A chemical solution is resting at a temperature of 73 degrees Celsius. The solution is placed in a freezer which is at a constant temperature of -15 degrees Celsius to crystallize. If it takes 30 minutes for the temperature of the mixture to drop to 25 degrees Celsius, how much longer will it take for the mixture to reach 0 degrees Celsius?
 A: The  Newton's law of cooling  states the following ODE for temperature changes:
$$\frac{dT}{dt} \propto T-T_a$$
or$$\frac{dT}{dt}=-k(T-T_a)\ \ \ \ \ \ldots (i)$$
(Here the use of minus sign is due to the fact that the temperature is decreasing.)
In this ODE, $t$ is the time, $T(t)$ is the temperature at time $t$ and $T_a$ is the temperature of the object's environment. In your case $T_a$ is the temperature of the freezer i.e $T_a=-15$. The Initial Value Conditions for your problem are
$$T(0)=73,\ T(30)=25$$
The solution of the ODE $(i)$ is
$$T=T_a+ce^{-kt}$$
$$T=-15+ce^{-kt}$$
By using $T(0)=73$, we get 
$$c=88$$
So
$$T=88e^{-kt}-15$$
Putting $T(30)=25$,
$$e^{-30k}=\frac{40}{88}$$
$$k=\frac{\ln\frac{40}{88}}{-30}\approx0.02628$$
So $$T=88e^{-0.02628t}-15$$
Now we have to fing the value of $t=t_\circ$ such that $T(t_\circ)=0$. So
$$88e^{-0.02628t_\circ}-15=0$$
$$t_\circ=\frac{\ln\frac{15}{88}}{-0.02628}\approx67.32$$
So almost 67 minutes after the initial time ($t=0$), the solution has temperature $0$degrees.
