- [Instructor] Welcome to a lesson
on Newton's Law of Cooling,
an application of first order differential equations.
The temperature T of a cooling object
drops at a rate that is proportional
to the difference T minus C,
again where T is the temperature of the cooling object
and C is the constant temperature
of the surrounding environment.
We can express this is a DT/DT equals
negative K times T minus C.
This is telling us the change in temperature
with respect to time of the cooling object
is equal to the opposite of a constant
times the quantity T minus C,
again where T is the temperature of the cooling object
and C is the temperature of the surrounding environment.
We will solve this differential equation to find T of T,
but the only function that satisfies
this differential equation or the general solution to this
would be T of T equals A times E
raised to the power of KT plus C
and we'll discover this in our first example.
A cup of coffee at 190 degrees Fahrenheit
is left in a room of 70 degrees Fahrenheit.
At time T equals zero, the coffee is cooling
at 15 degrees Fahrenheit per minute.
So using Newton's Law of Cooling given here below,
we need to be able to recognize that DT/DT,
the change in the temperature with respect to time
is equal to negative 15 degrees or negative 15
at T equals zero.
And the coffee starts at 190 degrees.
Therefore, capital T is equal to 190
and capital C or big C is the temperature of the room,
which is 70 degrees.
So using this information and Newton's Law of Cooling,
our first step is to find K,
the constant of proportionality.
So we want to solve the equation
negative 15 is equal to negative K
times the quantity T minus C, which would be 190 minus 70.
So here we have negative 15 is equal to
this is going to be negative 120K.
Divide both sides by negative 120.
So we have K is equal to this simplifies nicely to 1/8
or 0.125.
Now that we know K,
we're going to substitute this value
into Newton's Law of Cooling
to solve this differential equation.
So now we know DT/DT is equal to negative 0.125
times the quantity T minus C.
We know the temperature of the room will not change.
So we have T minus 70.
And we also know that big T of zero,
the starting temperature of the coffee is 190.
So now we can solve this as an initial value problem
and we can solve this using separation of variables.
So for the first step,
let's write this in differential form.
We have D of big T is equal to negative 0.125
times the quantity T minus 70 DT
and because we want big T on the left side of this equation,
we're going to multiply by the reciprocal of T minus 70,
which would be one over T minus 70.
So we have one over T minus 70 D big T
is equal to negative 0.125 D little T
and now from here we can integrate
both sides of the equation.
Our antiderivative here is going to be natural log
aptitude value of capital T minus 70,
but we know T is never going to be less than 70,
the temperature of the room.
So we can leave the aptitude value off.
We'll have natural log of T minus 70 or big T minus 70
equals negative 0.125 times little T
plus our constant of integration
and our goal is to solve this for big T.
So if this equation is true,
then E raised to the power on the left side of the equation
must equal E raised to the power on the right side.
So the left side simplifies nicely to big T minus 70
equals on the right side.
Notice how we're adding these exponents.
This is the same as E raised to the power of negative 0.125T
times E to the C.
Well if C is the constant,
E to the power of C is also a constant.
So from here, we're going to let E to the C
just equal a constant A.
So we can write the right side as
A times E to the negative 0.125T
and to solve for T, we're going to go ahead
and add 70 to both sides.
So we'll have plus 70.
And because this is a function of little T,
we have big T of little T is equal to this right side.
So now we can take this general solution
and use the initial condition big T of zero equals 190
to find the value of A and find our particular solution
and let's do that on the next slide.
So if big T of zero is equal to 190, then we have A
times E to the zero power, which is just one,
plus 70 equals 190.
So we have A plus 70 equals 190 and so we have A equals,
it's attracting 70 on both sides, 120.
So our particular solution
for the function that models the cooling of this coffee
is big T of T
equals again A is 120
E to the negative 0.125T plus 70.
And now we can use this function
to answer the last question,
which was how long will it take
for the temperature to reach 143 degrees?
So we'll set big T of T equal to 143
and then solve for little T.
So we'll subtract 70 on both sides.
So we'll have 73 equals 120 times E to the negative 0.125T.
Divide both sides by 120.
And now to solve for T, we'll take the natural log
of both sides of the equation
and then we can apply the power property of logarithms here
and move this exponent
and move this exponent to the front
and remember natural log E is equal to one.
So this gives us natural log of 73
divided by 120 equals negative 0.125T.
So we'll divide both sides by negative 0.125
to find the approximate value of T.
We'll need the calculator for this.
So we have natural log of 73 divided by 120
divided by negative .125.
So we'll say the time in minutes is approximately 3.98.
Okay, that's going to do it for this video.
I hope you found this helpful.