Newton's Law of Cooling // Separable ODE Example
Summary
TLDRIn this educational video, the host explores Newton's Law of Cooling by using a practical example of cooling tea. After setting the ambient temperature at 20°C (68°F), they record the tea's initial temperature at 71°C (161°F) and measure its temperature drop over time. By applying a separable differential equation, they model the cooling process, determine constants, and solve for the temperature function. The model's prediction is compared with experimental data, highlighting the approximation nature of models in real-world scenarios. The video concludes with a discussion on the validity and usefulness of the model.
Takeaways
- 📚 The video discusses Newton's Law of Cooling, a mathematical model that describes how the temperature of an object changes over time when it is not in thermal equilibrium with its surroundings.
- ☕ The presenter uses a cup of tea to demonstrate the concept, noting the ambient temperature of their house is 20°C or 68°F, which is the temperature they will compare against.
- 🔍 Initial data is collected by recording the tea's temperature at the start (161°F) and after two minutes (153.7°F), showing the tea is cooling over time.
- 📉 The rate of temperature change is modeled as proportional to the difference between the object's temperature and the ambient temperature, leading to a differential equation.
- 🔑 The differential equation is separable, allowing the variables to be separated and integrated on both sides to find a general solution for the temperature over time.
- ✅ The general solution is simplified using logarithms and exponential functions, resulting in a formula that includes constants a, d, and k, which are determined from the initial conditions and measurements.
- 🔍 The constant d is found to be 93 by using the initial temperature of the tea at time t=0.
- 🔢 The constant k is determined by taking a second temperature measurement at t=2 minutes, which allows for solving the equation involving k through logarithmic calculations, resulting in k=0.0409.
- 📉 The final temperature function is derived, showing the temperature of the tea as a function of time, which is a combination of the ambient temperature and the cooling effect described by the constants d and k.
- 📈 An additional data point is collected at t=5 minutes to test the model's accuracy, with the tea's temperature recorded as 142.7°F.
- 🤔 The model's prediction for t=7 minutes is calculated to be 137.8°F, which is close but not exactly the same as the experimental value of 142.7°F, indicating a reasonable but not perfect fit.
- 💭 The video concludes by emphasizing that models are approximations and are not expected to be perfectly accurate, but are useful for understanding and predicting phenomena like Newton's Law of Cooling.
Q & A
What is Newton's Law of Cooling?
-Newton's Law of Cooling is a principle that describes how the rate of cooling of an object is proportional to the difference between its own temperature and the ambient temperature of its surroundings.
What is the ambient temperature of the house mentioned in the script?
-The ambient temperature of the house is set at 20 degrees Celsius, which is equivalent to 68 degrees Fahrenheit.
What is the initial temperature of the tea mentioned in the video?
-The initial temperature of the tea is approximately 161 degrees Fahrenheit.
Why does the rate of temperature change depend on the difference between the object's temperature and the ambient temperature?
-The rate of temperature change depends on this difference because when the object's temperature is significantly higher than the ambient temperature, the rate of cooling is faster, and vice versa.
What type of differential equation is used to model the cooling process in the video?
-A separable differential equation is used to model the cooling process, which allows for the separation of variables to solve the equation.
How does the script differentiate between the temperature of the tea (T) and the ambient temperature (a)?
-The script denotes the temperature of the tea at time t as T(t) and the ambient temperature as a constant 'a', which is 68 degrees Fahrenheit.
What is the significance of the second temperature measurement taken at t = 2 minutes?
-The second measurement helps to determine the value of the constant 'k' in the differential equation, which represents the rate of cooling.
How is the constant 'd' in the model determined using the initial condition?
-The constant 'd' is determined by plugging in the initial condition (T(0) = 161 degrees Fahrenheit) into the model and solving for 'd', which turns out to be 93.
What is the process to find the value of 'k' in the model?
-The value of 'k' is found by using a non-zero time point (t = 2 minutes), comparing the model's prediction with the measured temperature, and then solving the resulting equation using logarithms.
How does the script evaluate the accuracy of the model?
-The script evaluates the model's accuracy by comparing its prediction at t = 7 minutes (137.8 degrees Fahrenheit) with the actual measured temperature (142.7 degrees Fahrenheit), noting a reasonable but not perfect match.
What is the final form of the temperature function as a function of time 't' according to the model?
-The final form of the temperature function is T(t) = a + d * e^(-kt), where 'a' is the ambient temperature, 'd' is the initial difference between the tea's temperature and the ambient temperature, and 'k' is the cooling constant.
Outlines
🔍 Introduction to Newton's Law of Cooling
The video begins with the presenter setting up an experiment to explore Newton's Law of Cooling, using a cup of tea to demonstrate the concept. The ambient temperature of the room is noted as 20 degrees Celsius, which is converted to 68 degrees Fahrenheit to match the scale of the thermometer used. The initial temperature of the tea is recorded at 161 degrees Fahrenheit. The presenter then introduces the idea of using a differential equation to model the rate of temperature change over time, explaining that differential equations are useful for describing rates of change. The rate of cooling is proposed to be proportional to the difference between the object's temperature and the ambient temperature, leading to the formulation of a differential equation to model this scenario.
📚 Solving the Differential Equation
The presenter proceeds to solve the differential equation derived from the cooling scenario. The equation is separable, allowing the variables to be separated and integrated on both sides. The integration results in a natural logarithm on the left and a linear function on the right. The presenter then isolates the variable 't' and uses exponential functions to simplify the equation further. Constants 'a', 'd', and 'k' are introduced, with 'a' being the ambient temperature, 'd' representing the initial temperature difference, and 'k' as a constant of proportionality. The presenter uses initial conditions and a second temperature measurement at two minutes to determine the values of 'd' and 'k', resulting in a specific function that models the temperature of the tea over time.
📉 Evaluating the Model's Accuracy
After establishing the mathematical model, the presenter tests its accuracy by comparing its predictions with an additional data point collected five minutes into the experiment. The model predicts a temperature of 137.8 degrees Fahrenheit at seven minutes, which is close but not exactly the same as the measured temperature of 142.7 degrees Fahrenheit. The presenter acknowledges the discrepancy and discusses the limitations of models, emphasizing that they are approximations and not exact representations of reality. The video concludes with a reflection on the model's validity and an invitation for viewers to consider the model's effectiveness and potential areas for improvement.
Mindmap
Keywords
💡Newton's Law of Cooling
💡Ambient Temperature
💡Differential Equations
💡Separable Differential Equation
💡Rate of Change
💡Proportionality
💡Exponential Growth
💡Constant of Proportionality
💡Initial Condition
💡Logarithm and Exponential Functions
💡Model Validation
Highlights
Introduction of Newton's Law of Cooling in the context of cooling a cup of tea.
Ambient temperature set at 20 degrees Celsius, converted to 68 degrees Fahrenheit.
Initial temperature of the tea recorded at 161 degrees Fahrenheit.
Use of a differential equation to model the rate of temperature change over time.
Differentiation between the rate of change in exponential growth and Newton's Law of Cooling.
Proportionality of the temperature change to the difference between the tea's temperature and ambient temperature.
Collection of data at two-minute intervals to observe temperature change.
Conversion of the differential equation into a separable form for easier solving.
Integration of the separable differential equation to find a general solution.
Use of initial conditions to determine the constant 'd' in the model.
Second temperature measurement at t=2 minutes to find the constant 'k'.
Calculation of the constant 'k' using a logarithmic approach.
Final temperature function derived as a function of time 't'.
Discussion on the practicality and accuracy of the model compared to experimental data.
Collection of an additional data point at t=5 minutes to test the model's prediction.
Comparison of the model's prediction with the experimental data, noting a reasonable match.
Reflection on the model's validity and the concept that models are approximations, not absolute truths.
Encouragement for viewers to engage with the content and leave comments for further discussion.
Transcripts
in this video we're going to talk about
newton's law of cooling
but before we do that i'm going to need
something to cool and i also really need
some caffeine so i think we can solve
both those problems
i'll also note that the ambient
temperature of my house is set at 20
degrees celsius
so before i even get into the video i'm
just going to record the fact that that
ambient temperature i'm going to write
that down as a was
20 degrees celsius and i'm actually
going to convert that to being
68 degrees fahrenheit just because the
thermometer you're going to see in a
moment
is going to be in fahrenheit as well
all right so i've got my tea but i now
want to figure out
exactly how hot is it and that's looking
about 161 degrees
so i'll write that down and i have that
the temperature which i'll denote by t
at time t equal to zero was 161 degrees
fahrenheit
and i'm actually going to collect one
more piece of data in two minutes i'll
have a temperature
at time t equal to two and i'm going to
come and fill that in
when i haven't all right so now i want
to model this scenario and we're going
to use
a separable differential equation indeed
this video is part of my entire
playlist on differential equations the
link to that playlist
as well as the free and open source
textbook that accompanies it is down in
the description
so why do i want to use a differential
equation to model this well
differential equations are great when
you can say something about the
rate of change and often the rate of
change of a quantity
is easier to say something about than
the quantity itself
so what kind of differential equation
should i write i'm going to try to study
the change in the temperature with
respect to time
the change in capital t is temperature
and lowercase t
is time i know that's just annoying and
the idea is
this is going to be proportional so i'll
write some constant of proportionality
to something but what should that
something be well when we were studying
exponential growth like for example the
spread of a pandemic you'd say
the rate of change is proportional to
how many people err infected so the rate
of change was proportional just to the
number to the quantity of the
variable that we were studying at that
time but now that doesn't seem quite as
reasonable
for example if the temperature in thy
cup is very close to
the ambient temperature i wouldn't
expect a very rapid change
but if the temperature was way way way
above the ambient temperature you might
expect a rapid
change so indeed i'm going to say this
is proportional
to the difference between the
temperature of my mug between my cup of
tea
and the ambient temperature so i will
model it as
k times the temperature of my mug
minus the ambient temperature and then
is this plus or minus okay
t is bigger than a because the
temperature of my cup is hotter than my
ambient temperature
so that's positive and then when it's
hotter it should be cooling the rate of
change should be going down it should be
a negative
so i'll put a negative there that is my
model for this scenario
all right so two minutes is up now let's
try measuring it one more time
and looks like we're at 153.7
just a little over two minutes later
okay so i'll just record that
observation this was about 153
degrees fahrenheit that's what happened
at time t equal to two
set timer for five minutes okay so let's
go and solve this
the first thing to observe is that this
is a so-called separable differential
equation
there's a portion on the right here that
only depends on the temperature
and there's also a portion here well in
fact there's no portion that depends on
time but
i'll say the portion that depends on
time is just this constant function
the minus k and so what i do is i
separate my variables
on the left hand side i'll write
everything to do with the temperature
so that had a dt and then i'll also have
the 1
over t minus a and on the right hand
side i'm going to have my minus k
and my dt so i completely separated the
lowercase t and the
uppercase t and then i'm going to do an
integral on both sides
and the details of this methodology of
separation of variables we've covered in
the previous video
in this course now it's an integration
problem
on the left this is going to be equal to
the logarithm
of t minus a i don't have to worry about
absolute values at all
because well it's a positive quantity on
the
right hand side i have my negative k
becomes
k t and then i cannot forget that
whenever you integrate indefinitely you
have this constant of integration so i
put my plus c down
okay so that is a solution but i can do
better i can solve for big t
i have a logarithm on both sides so let
me take e to the power of the logarithm
and e
to the power of minus kt plus c
on the left i observe that logarithm and
exponential are inverse functions of
each other
and thus these are going to cancel and
become t minus a
on the right hand side i have the e to
the minus k
t but how should i deal with that plus c
the plus c is it an exponent
so it can come out as e to the c
it would then be a multiplicative
constant the multiplicative constant e
to the c
but c is just some constant e to the c
is just some constant why don't i
re-label it and call it a new constant d
just a little it just looks a little bit
simpler this way
okay so i have an a i need to figure out
i have a d i need to figure out and i
have a k
i need to figure out three different
constants one of them i already know
one of them was the ambient temperature
and that was that 68 degrees fahrenheit
that we computed
to figure out the d i could plug in the
initial condition that we've seen at the
beginning that
t of 0 was 161 degrees fahrenheit
so let me try that t of 0 was 161
minus 68 is equal to d times e to the
k times zero i don't know what k is but
doesn't matter it's multiplied by zero
and e to the zero in fact i can even
erase it because it's just equal to one
so now i know that this d here is equal
to well
93. okay so i've got the a i've got the
d
but what about the k well the k was the
reason i took the second measurement
where i took the temperature at the
value of 2
and got this 153 the issue is that
because the k was multiplied by t
if you just plug in t equal to zero
you'll never figure out what the k was
so i need to have a different point a
non-zero point well let's now try to do
that
if this first computation was done at t
equal to zero i'll now do a computation
at t
equal to two what do i get on the left
is 153.7
minus the 68 and then on the right i now
know the value of d so i can put it in
the 93 and then it's going to be e
to the minus k times the value of 2.
so that's 85.7 on the
left but how do i even get to the k i
have to do some sort of logarithm
so to solve this i'm going to say that
85.7
divided out in fact by the 93 is equal
to e to the minus k times 2. then i'll
take my logarithm so
logarithm of 85.7 these are all such
funny numbers divided by 93
is equal to minus k times two
and in fact i'll take that two out from
the right hand side and divide it out on
the left
that is going to be my formula this is
much too complicated for me to do in my
head so i'm going to go to the
calculator
and the calculator tells me that k is
equal to
0.0409
i suppose that's enough decimal places
and so what's our final answer well we
have our
description of our problem here i have
my
a on the left i'm actually going to move
it over to the right hand side and i'm
going to get my
final answer which is the temperature
function as a function of t
i'll make it explicit is equal to the a
first a got moved to the other side so
68
plus the value of the d when the value
of the d
was 93 93 e to the negative
0.0409 times
t note by the way that this k is not
some universal property
it's about the geometry and the
materials
and the surface area of the mug of t
that i have it's for
my scenario i get this value of k you
have a different scenario you'll have a
different value of k
and so we have this model and that can
seem actually quite nice this is sort of
our
solution but how good is it so i want to
collect
one more data point and i did it five
minutes
later and now i'm all the way down to
142.7
so this new data point thus is t of 7
is equal to 142.7
okay so that's the experimental
doesn't mean my model is going to match
it at all so now we're going to get the
moment of truth
what happens if we use the model the
solution to that that we have come up
with
so now the t of 7 according to the model
is
68 plus 93
e to the negative 0.0409
times 7 moment of truth let's type into
the calculator
and this apparently is equal to 137.8
and so that is the model's prediction in
comparison
to the 142.7 which was our experimental
so this result is reasonable maybe not
excellent
we've definitely got a little bit of a
difference about five degrees here but
not completely terrible either
so then there's a bit of a question of
well do we like this result do we not
like this result
there might be for example some
experimental improvements we could have
done like recording the data
a little bit more accurately i was off
by at least 30 seconds
on one of my time measurements
but ultimately this leaves us with the
question of do we like this model or do
we not like the model
remember a model is not right or wrong i
mean no model you write down will ever
capture every single interaction down to
the quantum mechanical level for example
it's going to approximate it in some
level and that's what we're doing with
newton's law here of cooling
all right so i hope you enjoyed this
video if you did please give it a like
for the youtube algorithm everyone needs
to learn differential equations
i think so at least if you have any
questions leave them down in the
comments below
and we'll be doing some more math in the
next video
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