Welcome to an application problem
of Newton's law of cooling.
That results in the initial value problem
that can be solved using
the technique of separation of variables.
To begin,
Bob made a cup of coffee
and Bob likes to drink coffee
once it reaches 60 degrees Celsius and will not burn him.
Initially at time t equals zero minutes.
Bob measured the temperature
and the coffee was 89 degrees Celsius.
One minute later,
Bob measured the coffee again
and it was 85 degrees.
The temperature of the room,
the ambient temperature is 22 degrees.
When should Bob start drinking?
Let big T be the temperature of the coffee in degree Celsius
and let A be the ambient temperature or room temperature.
Also in degrees Celsius.
Newton's law of cooling states that
the rate at which the temperature of the coffee is changing
is proportional to the difference between
the ambient temperature
and the temperature of the coffee.
That is d big T dt
meaning the change in the temperature of the coffee
with respect to the time and minutes
is equal to some constant k
times the difference of A and big T
where A is the ambient temperature
and big T is a temperature of the coffee.
Before we go on,
let's record all the given information
or the initial conditions.
To begin at time zero,
the temperature is 89 degrees Celsius
which means big T of zero equals 89.
After one minute, the temperature is 85 degrees
which indicates big T of one equals 85.
And since the ambient or room temperature is 22 degrees
we know A is equal to 22.
To answer the question,
we want to find the time little t
such that big T of little t
is equal to 60 degrees Celsius.
And that's we're determining the particular solution
to the differential equation.
To begin, we know big A is equal to 22.
Let's go ahead and replace A with 22 giving us
d big T dt equals K times the quantity 22 minus T.
If you look at the textbook,
they do leave A in the equation,
but I'm going to go ahead and replace A with 22 now.
Because we're going to end up solving for big T,
it'll be helpful if big T is positive
and therefore let's factor out negative k from the right
giving us d big T dt is equal to negative k
times the quantity of T minus 22.
And now we solve the differential equation
using the technique of separation of variables.
We need the quantity T minus 22 on the left
and therefore we multiply both sides of the equation
by one divided by the quantity T minus 22.
And then we write the differential equation
and differential form
where we can think of multiplying
both sides of the equation by dt
which gives us one divided by the quantity
big T minus 22 dT equals negative k times dt.
Now we integrate both sides of the equation.
The in depth integral of one
divided by the quantity big T minus 22
with respect to big T
is equal to natural log
absolute value of big T minus 22,
plus a constant,
but we'll include the constant on the right.
On the right,
the in-depth integral of negative k with respect to t
is equal to negative kt plus C.
And now we need to solve the equation for big T.
We can do this two ways.
We can write the log equation as an exponential equation
or we can exponentiate both sides of the equation
with the base of e.
Either case,
natural log of the opposite value of big T minus 22
equals negative kt plus C
is equivalent to
the quantity big T minus 22 equals e
raised the power of negative kt plus C.
Notice I drop the absolute value around big T minus 22
because the exponential on the right is always positive
and because the exponent on the right is a sum
we can write this as e to the power of negative kt
times e the power of C
or e to the power of C is just some constant.
Let's let this equal D
and therefore we can solve for big T
and get big T is equal to 22 plus D
times e to the power of negative kt.
And now that we have the general solution
we need to determine D and k
using the initial conditions.
Let's do this on the next slide.
Again, there's some work here from the textbook.
I'm going to go ahead and show some additional work.
Let's first determine the value of D
using the initial condition,
big T of zero equals 89.
To do this, we substitute zero for T and 89 for big T.
This gives us 89 equals 22 plus D
times e to the power of negative k times zero
weight of the zero is equal to one
and therefore solving for D
we simply subtract 22 on both sides
and we have D equals 67.
So now we know
big T is equal to 22 plus 67
times e to the power of negative kt.
And now we can use a second condition
of big T of one equals 85 to determine k.
To do this, we substitute one for t and 85 for big T.
This gives us 85 is equal to 22
plus 67 times e to the power of negative k times one,
which is just negative k.
And now we solve for k.
First, we subtract 22 on both sides
and then we divide both sides by 67.
Next, we take the natural log
of both sides of the equation
to solve for k.
On the right, natural log of e to the power of negative k,
applying the power property of logs,
is equal to a negative k times natural log e,
but natural log e is one,
giving us negative k equals the natural log of 63/67.
Next, we divide both sides by negative one
and go with a calculator.
Negative natural log of 63/67
is approximately 0.0616.
So now we have that particular solution.
We have big T is equal to 22 plus 67 times e
raised to the power of negative 0.0616t.
And now we can finally answer the question,
how long will take for the coffee to reach
60 degrees Celsius
by solving the equation big T of t equals 60.
Meaning we substitute 60 for big T
and solve for time little t.
Let's do this on the next slide.
Here's the equation we want to solve.
We first subtract 22 on both sides.
Next we divide both sides by 67.
Take the natural log of both sides of the equation.
The right side simplifies nicely to negative 0.0616t.
To solve for t, we divide both sides by negative 0.0616.
Going to the calculator,
we have t is approximately 9.206 minutes.
This means,
Bob's coffee will reach 60 degrees Celsius
in approximately 9.2 minutes,
and therefore we should wait 9.2 minutes
before he drinks his coffee.
Before we go,
let's also look at this graphically.
In both graphs, the dark blue function
is a temperature function big T of t.
On the left,
horizontal lines are drawn at temperatures,
60 degrees in pink,
85 degrees in brown,
and 89 degrees in green.
The vertical lines are done at t equals one
and t equals 9.21.
That's here and here.
Notice that the temperature of the coffee hits 85 degrees
at t equals one minute here,
and 60 degrees at approximately 9.21 minutes
which is this intersection point here.
On the right, the graph is over a long period of time
with a horizontal line
at the ambient temperature, 22 degrees Celsius.
I hope you found this helpful.