Newton's Law of Cooling // Separable ODE Example

Summary

In 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

00:00

🔍 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.

05:01

📚 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.

10:01

📉 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.

Keywords

💡Newton's Law of Cooling

Newton's Law of Cooling is a principle in physics that describes the rate at which an object cools down and approaches the ambient temperature. In the video, this law is the central theme, as the presenter uses it to model the cooling process of a cup of tea. The script illustrates the law by showing how the temperature of the tea decreases over time, comparing it to the ambient temperature of the room.

💡Ambient Temperature

Ambient temperature refers to the temperature of the surrounding environment. In the context of the video, it is the initial temperature of the room, set at 20 degrees Celsius or 68 degrees Fahrenheit. The script uses this term to establish a baseline against which the cooling of the tea is measured, highlighting the difference between the tea's temperature and the ambient temperature as a key factor in the cooling process.

💡Differential Equations

Differential equations are mathematical equations that involve rates of change, making them ideal for modeling situations where something is changing over time. The video script discusses using a differential equation to model the cooling of the tea, emphasizing their utility in describing the rate of temperature change rather than the temperature itself. The script also mentions a playlist and a textbook related to differential equations, indicating their broader relevance in mathematical modeling.

💡Separable Differential Equation

A separable differential equation is a type of differential equation where the variables can be separated on either side of the equation, simplifying the process of solving it. The script describes the process of separating variables in the context of modeling the tea's cooling, which involves integrating both sides of the equation to solve for the temperature as a function of time.

💡Rate of Change

The rate of change refers to the speed at which a quantity changes over time. In the script, the rate of change is used to describe how quickly the temperature of the tea decreases as it cools. The differential equation in the video is based on the principle that the rate of temperature change is proportional to the difference between the tea's temperature and the ambient temperature.

💡Proportionality

Proportionality is a mathematical concept where two quantities are related such that a change in one results in a proportional change in the other. The script uses the idea of proportionality to establish the relationship between the rate of temperature change and the difference between the tea's temperature and the ambient temperature, which is a key aspect of the differential equation model.

💡Exponential Growth

Exponential growth is a pattern of increase where a quantity grows at a rate proportional to its current value. The script contrasts exponential growth with the cooling process, using the spread of a pandemic as an example where the rate of change is proportional to the current number of infected people. This comparison helps to illustrate the different types of relationships that can be modeled using differential equations.

💡Constant of Proportionality

The constant of proportionality, often denoted as 'k' in the script, is a factor that relates two proportional quantities in a mathematical equation. In the context of the video, 'k' is used in the differential equation to represent the rate at which the tea cools, which is a property specific to the tea, the mug, and the environment.

💡Initial Condition

An initial condition is a value or set of values that apply at the start of a process. The script identifies the initial temperature of the tea at time zero as an initial condition, which is essential for solving the differential equation and determining the constants in the model. The initial condition helps to establish the starting point for the cooling process.

💡Logarithm and Exponential Functions

Logarithm and exponential functions are inverse mathematical operations used to solve equations involving growth or decay. In the script, the presenter uses logarithms to solve the differential equation for the tea's cooling, and exponential functions to express the solution. These functions are key to deriving the formula that describes how the temperature of the tea changes over time.

💡Model Validation

Model validation is the process of comparing a model's predictions with experimental data to assess its accuracy. The script describes the presenter's attempt to validate the cooling model by comparing its predictions with actual temperature measurements taken at different times. The comparison shows a reasonable match, indicating that the model is a good approximation of the cooling process, although it is not perfect.

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