Newton's Law of Cooling Video Lecture - Class 11

- [Voiceover] Let's think
about another scenario
that we can model with the
differential equations.
This is a scenario where we take an object
that is hotter or cooler than
the ambient room temperature,
and we want to model how
fast it cools or heats up.
And the way that we'll think about it is
the way that Newton thought about it.
And it is described as
Newton's Law of Cooling.
And in a lot of ways, it's common sense.
It states that the rate
of change of temperature
should be proportional
to the difference between
the temperature of the object
and the ambient temperature.
So let me write that
in mathematical terms.
So Newton's Law of Cooling tells us,
that the rate of change of temperature,
I'll use that with a capital T,
with respect to time, lower case t,
should be proportional to the difference
between the temperature of the object
and the ambient temperature.
So that is a mathematical
description of it.
And once again, it's common sense.
If something is much, much hotter
than the ambient temperature,
the rate of change
should be pretty steep, it should be
declining in temperature quickly.
If something is much, much cooler,
it should be increasing
in temperature quickly.
And if something is
close, if these two things
are pretty close, well
maybe this rate of change
shouldn't be so big.
Now I know one thing that you're thinking.
You're like, okay, if
the temperature is hotter
than the ambient temperature,
then I should be cooling.
My temperature should be decreasing.
And a decreasing temperature would imply
a negative instantaneous change.
So how will this be a negative value
in the case where our
temperature of our object is
greater than our ambient temperature?
And the way that that would happen is,
you would have to have a negative k.
If you don't like thinking
in terms of a negative k,
you can just put a
negative right over here
and now you would have a positive k.
Now it makes sense.
If our thing is hotter, if
it has a higher temperature
than the ambient temperature,
so this is a positive,
then our rate of change will be negative,
will be getting cooler.
If it was the other way around, if our
temperature of our
object is cooler than our
ambient temperature, then this thing
is going to be a negative,
and then the negative of that
is going to be a positive,
we're assuming a positive k,
and our temperature will be increasing.
So hopefully, this makes
some intuitive sense.
And our constant k could depend on
the specific heat of the
object, how much surface area
is exposed to it, or whatever else.
But now I'm given this,
let's see if we can solve
this differential equation
for a general solution.
And I encourage you to pause
this video and do that,
and I will give you a clue.
This is a separable differential equation.
So I assume you've had a go at it,
so let's now work through it together.
So, we just have to
algebraically manipulate this
so all my Ts and dTs are on one side.
I should say, so all my capital
Ts and dTs are on one side,
this is going to be a
little bit more confusing
because I have a capital
T and a lower case t.
Capital T for temperature,
lower case t for time.
But hopefully we'll be
able to work through it.
And then I'm going to have
all my time differentials
and time variables on the other side.
So one thing I could is
I could divide both sides
by T minus ambient
temperature, minus T sub a.
Remember this is just
going to be a constant
based on what our ambient temperature is.
We're going to assume
our ambient temperature
doesn't change as a function of time,
it's just such a big
room that our cup of tea
is not going to actually warm up the room.
So that's just one of these assumptions
that we're going to make.
So if we do that, if we
divide both sides by this,
we are going to have...
So I'm going to divide both sides,
I'm going to do this in a new color.
If I divide both sides
by that, I get one over T
minus T sub a, and let
me multiply both sides
times the time differential.
So I'm going to have, that dT,
our temperature differential.
Times our temperature differential,
is going to be equal to
negative k times our time differential.
So once again, to separate the variables,
all I did was divide both sides by this,
and multiply both sides by that.
Now I can integrate both sides,
we've seen this show before.
So I can integrate both sides.
And the integral of this is going to be
the natural log of the absolute value
of what we have in the denominator.
And you can do u substitution if you want.
If we said u is equal to T minus T sub a,
then du is just going to be one dT,
and so this is essentially,
you could say the integral
of one over u du, and so
it would be the natural log
of the absolute value of u,
and this right over here is u.
So this is the natural
log of the absolute value
of T minus T sub a, is equal to,
and once again I could
put a constant here,
but I'm going to end up with a constant
on the right hand side too so
I'm just going to merge them
into the constant on the right hand side.
So that is going to be equal to, now here,
this is going to be negative kt,
and once again we have plus C.
And now we can raise e
to both of these powers,
or another way of interpreting this
is if e to this thing is
going to be the same as that.
So we can write this
as, the absolute value,
let me do that in that same blue color.
We can write this as the
absolute value of T minus T sub a
is equal to e, something about e I always
think of the color green.
e to the negative kt plus C.
This of course is the same thing as,
this is equal to e to the negative kt,
we've done this multiple times before.
Negative kt times e to the C power.
And we could just call this
another arbitrary constant.
If we called this C1,
then we could just call
this whole thing C.
So this we could say is
Ce to the negative kt.
So at least it's starting
to resemble what we did
when we were modelling population.
We'll see it's a little bit different.
Instead of just temperature
on this left hand side,
we have temperature minus
our ambient temperature.
And so, we can do a couple of things.
If, in a world, say we were
dealing with a hot cup of tea,
something that's hotter than
the ambient temperature.
So we could imagine a world where T is
greater than or equal to
our ambient temperature.
So that means this is hot, or it's hotter,
I guess we could say.
So if we're dealing with something
hotter than the ambient temperature,
then this absolute value
is going to be positive
or the thing inside the absolute value
is going to be positive.
So we don't need the absolute value.
Or the absolute value of it is going to be
the same thing as it.
And then we can just add
T sub a to both sides,
and then we would have our temperature,
and I can even write this
as a function of time,
is going to be equal to this business,
is going to be equal to Ce, let me do that
in that same color.
Ce to the negative kt plus T sub a.
All I did is I'm assuming that this
inside the absolute value
is going to be positive,
so the absolute value is not
going to change the value.
And I added T sub a to
both sides to get this.
So this right over here is going
to be our general solution,
in the case where we start
with something that is hotter
than the ambient room temperature.
And if we want to look at the
case where something is cooler
than the ambient room temperature,
so that's the situation,
let's say T is less than
our ambient room temperature.
Then the absolute value of
T, then this thing over here
is going to be negative, and
so the absolute value of it's
going to be the negative of that.
So then this up here
results in T sub a minus T,
that's going to be the same
thing as the absolute value,
it's going to be the
negative of the negative.
So then that is going to be
equal to e to the negative k
plus, actually let me just do it...
T sub a minus T is going to be equal to Ce
to the negative kt, so
this is equal to that.
I'm just assuming that
T is less than T sub a.
And so then, to solve for T,
you could add T to both sides
and subtract this from both sides.
You would have T as a function
of t is going to be equal to,
let's see, if this went onto that side
and this goes over here,
you would have T sub a
minus Ce to the negative kt.
Did I do that right?
So yep, that looks right.
So this is the situation where
you have something that is
cooler than the ambient temperature.
So this right over
here, based on the logic
of Newton's Law of Cooling,
these are the general solutions
to that differential equation.
In the next video we can actually apply it
to model how quickly something
might cool or heat up.