So we have this relationship for Newtons law of cooling, in this case a bowl of soup
$$\frac{dT}{dt} = -k(T – T_a)$$
where Ta is the temperature of the room and T is the temperature of the soup at that time and $T_a = -10$. (Both in celsius)
Given that I have 2 readings at 2 different times where $T_1 = 20$ and $T_2 = 10$ at times 9am and 11am.
We do not know when the soup started to cool.
But we know $$dT = 20 – 10 = 10$$ and $$dt = 11 – 9 = 2$$
And because we know that the change in temperature in respect to time is equal to $$ -k(T – T_a) $$
we can say that T is simply the change in temperature. And this equation to find k is valid
$$ \frac{10}{2} = -k((20 – 10) + 10)$$
Because the change in temperature from 9am to 11am is the same as the soups temperature after 2 hours since the temperature change is proportional to the temperature at t.
But this isn't correct. Can someone straighten my intuition please?
Best Answer
Your intuition is wrong because you are treating the infinitesimally small values $dT$ and $dt$ as regularly-sized values.
You say that $dt=2$, and there is some validity to that, but it is more accurate to say that $\Delta t=2$. The delta ($\Delta$) is the difference between the two values at different times. It is definitely wrong to write that $dT=10$: you definitely need to say that $\Delta T=10$.
The temperature $T$ does not change linearly, it is instead decreasing with an exponential decay toward the room temperature $T_a$. Your calculated value of $\Delta T/\Delta t=5$ is merely the average decay rate between those two times. That says nothing about the decay rate at any particular time and little about the decay between any two particular times. You cannot treat a non-linear process the same as a linear process. Your intuition would work fine for a linear process but fails for this non-linear one.
Here is the correct intuition for what would happen for the next two hours after 11:00 a.m., which works for this exponential-decay process. At 9:00 a.m. the temperature is $20$, which is $20-(-10)=30$ above the room temperature. AT 11:00 a.m. the temperature is $10$, which is $10-(-10)=20$ above the room temperature. So in two hours the temperature difference between the soup and the room dropped to $2/3$ its beginning value. Since this is an exponential-decay process, that means that in the next two hours the temperature difference will again drop by the proportion of $2/3$. The temperature difference at 11:00 a.m. is $20$ so the difference will be $20 \cdot 2/3 = 13+1/3$ at 1:00 p.m. Thus the actual temperature will be $13+1/3$ above the room temperature $-10$ which is $3+1/3$. In another two hours the temperature difference will be reduced to $2/3$ again, and so on.
So your intuition needs to look at the temperature differences between the soup and the room, not the soup temperatures themselves. And your intuition needs to look at the proportional rate of decrease of those temperature differences, not the absolute changes.
Of course, to find the temperature differences at time periods other than every two hours you would use logarithms and exponentials. That is what you see when you solve the differential equation you started with.