Suppose that a cup of soup cooled from 90 degrees C to 70 degrees C after 20 minutes in a room whose temperature was 20degreesC. Use Newton's law of cooling to answer the following questions.
(A) How much longer would it take the soup to cool to 45 degree C?
(B) Instead of being left to stand in the room, the cup of 90 degree C soup is put in the freezer whose temperature is - 15 degree C. How long will it take the soup to cool from 90 degree C to 45 degree C?

Newton's law of cooling is an empirical, simplified model that states that the instantaneous rate of cooling is proportional to the difference in temperature between an object and its environment.

we can write it as:

[tex]\cfrac{dT}{dt}=-k(T-T_o)[/tex]

Where [tex]T[/tex] is the body's temperature and [tex]T_o[/tex] is the environment's temperature. This equation is a first order differential equation and it can be solved quite simply, if we rearrange it as follows:

[tex]\cfrac{dT}{dt}+k\, T=k\, T_o\\\\e^{kt}\cfrac{dT}{dt}+k\,e^{kt}\, T=k\,e^{kt}\, T_o\\\\\cfrac{d(e^{kt}\, T)}{dt} = k\,e^{kt}\, T_o\\\\e^{kt}\, T = e^{kt}\, T_o+C\\\\T=T_o+C\, e^{-kt}[/tex]

We have found a general solution! We can see now clearly what happens as t grows, the temperature of the body tends exponentially towards the environment's temperature [tex]T_o[/tex]!

Now To solve this problem we need to find the values for our constants C and k. The constant C is related to the initial value of our temperature, while k doesn't change at all.

IF we plug in the values we got, we will arrive to:

[tex]T(0)=T_o+C\, e^{0}=20 + C = 90\\\\C=70\\\\T(20)=70=20+70\cdot e^{-k\, 20}\\\\\cfrac{50}{70}=e^{-k\, 20}\\\\\ln\left(\cfrac{50}{70}\right) =-k\, 20\\\\k=\cfrac{\ln\left(\cfrac{70}{50}\right)}{20} =0.01682[/tex]

Now we are in position to answer the questions. To answer A), we need to plug in [tex]T(t)=45[/tex] and find t:

[tex]45=20+70\cdot e^{-kt}\\\\\cfrac{25}{70}=e^{-kt}\\\\-\cfrac{\ln \cfrac{25}{70}}{k}=t=61.2[/tex]

Now, as the question is asking how much longer it is going to take, we have to substract the 20 minutes that have first elapsed, so we get [tex]\Delta t = 41.2[/tex]

Now to solve B, [tex]T_o = -15[/tex], Therefore C must also change:

[tex]T(0) = T_o +C =-15+C=90\\\\C=105[/tex]

Now we only have to repeat the last question's calculations, with different values for C and [tex]T_o[/tex]

We will then get:

[tex]t=33.26[/tex]