Linear Expansion of Solids, Volume Contraction of Liquids, Thermal Physics Problems

Summary

This educational video delves into the concept of thermal expansion, illustrating how objects change in size when subjected to temperature variations. It explains that solids expand with heat and contract with cold, with the change in length (ΔL) being proportional to the original length, temperature change, and a material-specific coefficient (α). The video uses practical examples, such as an aluminum bar and a lead plate, to demonstrate calculations for length and area changes due to temperature. It also covers volume expansion, introducing the coefficient of volume expansion (β) and its relation to linear expansion (α), with examples including brass and water. The video simplifies complex physics into accessible explanations, making it an informative resource for learners.

Takeaways

• 🔥 Thermal expansion is the process where objects increase in size when heated and decrease when cooled.
• 📏 The change in length (ΔL) of an object due to temperature change is directly proportional to its original length (L) and the temperature change (ΔT).
• 🌡️ The coefficient of linear expansion (α) is a material-dependent constant that relates the change in length to the temperature change.
• 📐 For a given material, the longer the object, the greater the change in length for the same temperature change.
• 🛠️ The formula for calculating the change in length is ΔL = α * L * ΔT, where units of length must be consistent.
• 📉 When the temperature decreases, ΔL is negative, indicating contraction of the object.
• 📏 The new length of an object can be found by adding the original length to the change in length (L_new = L + ΔL).
• 🔩 The area of an object expands or contracts when its length and width change due to temperature variations.
• 📐 The change in area (ΔA) can be approximated by ΔA ≈ 2 * L * W * α * ΔT for small temperature changes.
• 💧 Volume expansion occurs similarly to linear expansion, with the change in volume (ΔV) related to the original volume, the coefficient of volume expansion (β), and the temperature change.
• 📦 For solids, β is approximately three times α, simplifying the volume expansion calculation to ΔV = V * β * ΔT.

Q & A

What is thermal expansion?

Thermal expansion is the phenomenon where objects increase in size when heated and decrease in size when cooled.

How does the temperature change affect the length of a solid bar?

When the temperature of a solid bar is increased, it expands and gets longer; when the temperature is decreased, it contracts and gets shorter.

What is Delta L in the context of thermal expansion?

Delta L represents the change in length of an object due to a change in temperature.

What is the relationship between the change in length (Delta L) and the original length (L) of an object?

The change in length (Delta L) is proportional to the original length (L) of the object, meaning a longer object will experience a greater change in length for the same temperature change.

What is the coefficient of linear expansion (Alpha), and how does it relate to the material of an object?

The coefficient of linear expansion (Alpha) is a constant that depends on the type of material and determines how much an object's length changes with temperature.

How can you calculate the change in length of an aluminum bar when its temperature is increased?

You can calculate the change in length (Delta L) using the formula: Delta L = Alpha * L0 * Delta T, where Alpha is the coefficient of linear expansion, L0 is the original length, and Delta T is the change in temperature.

What is the new length of a bar after a temperature change, and how do you calculate it?

The new length of a bar after a temperature change is the original length plus the change in length (L_new = L0 + Delta L).

How does thermal expansion affect the area of a solid plate?

Thermal expansion affects the area of a solid plate by causing it to expand in all directions, increasing both its length and width, which results in an increased area.

What is the formula for calculating the change in area (Delta A) of a solid plate due to thermal expansion?

The formula for calculating the change in area (Delta A) is: Delta A ≈ l0 * w0 * (2 * Alpha * Delta T), where l0 and w0 are the original length and width, Alpha is the coefficient of linear expansion, and Delta T is the change in temperature.

How does the volume of an object change with temperature, and what is the coefficient of volume expansion (Beta)?

The volume of an object changes with temperature due to thermal expansion, and the coefficient of volume expansion (Beta) is used to calculate this change, typically being approximately three times the linear expansion coefficient (Alpha) for most solids.

What is the relationship between linear expansion and volume expansion in terms of their coefficients?

For most solids, the coefficient of volume expansion (Beta) is approximately three times the coefficient of linear expansion (Alpha), reflecting the three-dimensional nature of volume change compared to the one-dimensional change in length.

Outlines

00:00

🔥 Understanding Thermal Expansion

This paragraph introduces the concept of thermal expansion, explaining how objects change in size when their temperature is altered. It uses the example of a bar with an initial length 'L', which expands or contracts depending on temperature changes. The expansion or contraction is quantified by 'Delta L', which is directly proportional to the temperature change and the original length of the bar. An equation is provided to calculate the change in length, which involves the coefficient of linear expansion (Alpha), the original length, and the temperature change. The paragraph concludes with a practice problem involving an aluminum bar, demonstrating how to calculate the change in length when the temperature increases from 20°C to 75°C.

05:05

📏 Calculating Area Change in a Lead Plate

The second paragraph delves into the impact of thermal expansion on the area of a solid object, exemplified by a lead plate. It discusses how both the length and width of the plate will change when the temperature is raised, leading to an increase in area. The calculation involves determining the change in length (Delta L) and width (Delta W) using the coefficient of linear expansion and the change in temperature. The new area is then calculated by multiplying the new dimensions. A practice problem calculates the change in area and the new area of a lead plate when the temperature rises from 15°C to 95°C.

10:11

📐 Deriving the Formula for Area Change

This section explains the derivation of the formula for calculating the change in area of a solid object due to thermal expansion. It starts with the premise that both the length and width of an object expand when the temperature increases. The formula Delta a equals l0*w0*2*Alpha*Delta T is introduced, where l0 and w0 are the original dimensions, Alpha is the coefficient of linear expansion, and Delta T is the change in temperature. The paragraph also discusses the negligible impact of the term involving Alpha squared, leading to an approximation of the formula that focuses on the primary term.

15:11

💧 Volume Expansion in Solids and Liquids

The fourth paragraph explores volume expansion, applying the concept to both solids and liquids. It explains that when the temperature of a rectangular solid is increased, its volume changes according to the equation Delta V equals the coefficient of volume expansion (Beta) times the original volume times the change in temperature. The relationship between Beta and Alpha is discussed, noting that Beta is approximately three times Alpha for most solids. The paragraph includes a problem involving a brass rectangular solid and a cup of water, demonstrating how to calculate the change in volume for both an increase and a decrease in temperature.

20:13

📦 Calculating Volume Change Using Linear Expansion

This paragraph discusses how to calculate the change in volume using linear expansion coefficients. It explains that the change in volume (Delta V) can be approximated by multiplying the original volume (V) by three times the linear expansion coefficient (Alpha) and the change in temperature. The process involves expanding the expression (1 + Alpha*Delta T)^3 and simplifying it to 3*Alpha*Delta T, which is then related back to the coefficient of volume expansion (Beta). The paragraph concludes by showing that the equation for volume change due to temperature is essentially the original volume times Beta times the change in temperature.

25:15

🌡️ Final Thoughts on Thermal Expansion

The final paragraph wraps up the discussion on thermal expansion by summarizing the key points covered in the video. It reiterates the relationship between linear and volume expansion and thanks the viewers for watching. The paragraph serves as a conclusion, highlighting the importance of understanding how temperature changes affect the size of materials, which is crucial for various applications in science and engineering.

Keywords

💡Thermal Expansion

Thermal expansion is a physical phenomenon where the size of a material changes in response to a change in temperature. In the video, thermal expansion is the central theme, as it explains how objects expand when heated and contract when cooled. The script provides the formula for calculating the change in length (ΔL) due to thermal expansion, which is crucial for understanding how materials react to temperature changes.

💡Coefficient of Linear Expansion (Alpha)

The coefficient of linear expansion, denoted as Alpha (α), is a material property that quantifies the change in size of a material for a unit change in temperature. It is a key factor in the formula for thermal expansion and is specific to each material. The video uses the example of an aluminum bar to demonstrate how Alpha is used in calculations to determine the change in length when the temperature changes.

💡Delta L (ΔL)

Delta L (ΔL) represents the change in length of an object due to thermal expansion or contraction. The script explains that ΔL is directly proportional to both the original length of the object and the change in temperature. It is calculated using the formula ΔL = α * L0 * ΔT, where α is the coefficient of linear expansion, L0 is the original length, and ΔT is the change in temperature.

💡Temperature Change (ΔT)

Temperature change, denoted as ΔT, is the difference between the final and initial temperatures. In the context of the video, ΔT is a critical variable in the thermal expansion formula, as it determines the extent of expansion or contraction. The script uses ΔT in the example problems to calculate the change in length of a bar and the change in area of a plate when the temperature changes.

💡Length Contraction

Length contraction refers to the reduction in the size of an object when its temperature is decreased. The video script explains that when an object cools, it contracts, and this is represented by a negative ΔL in the thermal expansion formula. This concept is important for understanding how materials behave in cold conditions.

💡Coefficient of Volume Expansion (Beta)

The coefficient of volume expansion, or Beta (β), is analogous to Alpha but relates to the change in volume of a material per unit change in temperature. The video mentions that for most solids, Beta is approximately three times Alpha. This concept is used to calculate the change in volume of an object when its temperature changes.

💡Change in Area (ΔA)

The change in area (ΔA) is the difference between the final and initial areas of an object as it undergoes thermal expansion or contraction. The video script explains how to calculate ΔA using the formula ΔA ≈ l0 * w0 * 2 * α * ΔT, where l0 and w0 are the original length and width, respectively. This formula is used in the example of a lead plate to find the change in area when the temperature increases.

💡Change in Volume (ΔV)

Change in volume (ΔV) is the difference between the final and initial volumes of an object due to thermal expansion or contraction. The video uses the formula ΔV = V * β * ΔT to calculate ΔV, where V is the original volume, β is the coefficient of volume expansion, and ΔT is the temperature change. This concept is illustrated with examples involving a brass rectangular solid and water in a cup.

💡Original Length (L0)

Original length (L0) is the initial size of an object before any temperature change occurs. In the video, L0 is used as a reference point in the thermal expansion formula to calculate the change in length (ΔL). The script emphasizes that the change in length is proportional to the original length, highlighting its importance in determining the extent of expansion.

💡New Length

New length refers to the length of an object after it has undergone thermal expansion or contraction. The video script explains how to find the new length by adding the change in length (ΔL) to the original length (L0). This concept is crucial for understanding the final dimensions of an object after it has been subjected to temperature changes.

Highlights

Thermal expansion causes objects to change size with temperature fluctuations.

A solid object expands when heated and contracts when cooled.

The change in length (Delta L) is directly proportional to the temperature change.

The change in length is also proportional to the original length of the object.

The coefficient of linear expansion (Alpha) is a material-dependent constant.

The formula for change in length is Delta L = Alpha * L * Delta T.

An example problem involves calculating the change in length of an aluminum bar.

The new length of an object can be found by adding the original length to the change in length.

The area of a solid plate expands in both dimensions when the temperature increases.

The change in area can be approximated using the formula Delta A ≈ 2 * l0 * w0 * Alpha * Delta T.

Volume expansion is related to the coefficient of volume expansion (Beta), which is approximately 3 times Alpha for most solids.

The change in volume (Delta V) can be calculated using the formula Delta V = V * Beta * Delta T.

For small Alpha, terms involving Alpha squared and Alpha cubed can be neglected in volume expansion calculations.

An example demonstrates how to calculate the change in volume of a brass rectangular solid.

The volume of liquids also changes with temperature, as shown with an example of water in a cup.

The video concludes with a summary of how linear expansion relates to volume expansion.

Transcripts

in this video we're going to talk about

thermal

expansion what happens to objects when

they're heated or

cooled well let's find out let's say if

we have a

bar that has a length of

L and what's going to happen if we raise

the temperature whenever you increase

the

temperature a solid will expand it's

going to get B bigger and if you cool it

it's going to contract it will get

smaller so if we raise the

temperature this bar will

increase by an amount called Delta L

which represents the change in

length so as you increase the

temperature Delta L

increases if you decrease the

temperature Delta L will be negative it

will contract so the bar will be smaller

if you cool

in Now Delta L is not only proportional

to the temperature but it's also

proportional to the

length if you increase the length of the

rod the change in length will also

increase if you of course increase the

temperature

the equation that relates these two is

this equation the change in

length is equal

to Alpha which is the coefficient of

linear

expansion times the original length

times the change in

temperature

Alpha is a coefficient or a constant

that depends on the type of material

so the change in length depends on the

material how long the bar is the length

of the bar and also how much the

temperature changes by so let's work on

some practice

problems and aluminum bar is 1.25 M long

at 20°

C the coefficient of linear expansion is

25 * 10-

6 and that's supposed to be Celsius

min-1 if the temperature is increased to

75°

C how much will the length of the bar

change so we're looking for Delta

l so Delta L is equal to Alpha * l0 *

Delta

C the coefficient of linear expansion is

25 * 10 to Theus 6 and the units are

Celsius minus1 which is basically 1 over

Celsius

the original length is 1.25

M and the change in

temperature final minus initial 75 minus

20 the change in temperature is

positive 55° C because it increased by

55 going from 20 to

75 so notice that the unit Celsius

cancel so whatever unit l0 is in that

will be the unit for Delta l so l0 could

be in meters

centimeters it doesn't matter these two

will have the same

unit so in this problem since l0 is in

meters Delta L will be in

meters so now let's go ahead and

calculate the

answer so 25 * 10 - 6 time

1.25 *

55 is equal to about

.00

17 2 m so the length doesn't change much

so this is the answer uh to part A

that's how much the length of the bar

changes less than a centimeter so now

what is the new length of the bar at

this temperature to find the new length

we need to add the original length plus

the changing length

so it's going to be l0 plus Delta

l so that's 1.25 M plus

0. 172

M so this is about

1.25 17 but you can round that to 1.25 2

if you want so this is the new length of

the bar as you can see it increased uh

slightly let's work on this one a solid

plate of lead is 8 cm x 12 CM at 15° C

and we have the coefficient of linear

expansion it's 29 * 10 -

6 um 1/ Celsius now what is the change

in area and also the new area of the

lead plate if the temperature increases

to 95

so let's draw a

picture so the lead plate is 8 by

12 so this side is 8 cm and this side is

12

CM if we multiply 12 * 8 that's a

current area of

96 square

cm now it's important to understand that

once we raise the temperature we know

it's going to expand but it's going to

expand in all

directions so

if let me draw a new

picture if this side is

l0 then it's going to expand by Delta L

and if this side is w0 for the width

then it's going to expand this way

by Delta w

so not only will expand to the right but

it will also expand down as well so the

length and the width will

increase so let's calculate the change

in left first and then we'll calculate

the change in

width so Delta

L is equal to Alpha * l0 * Delta

C so Alpha is 29 * 10- 6 the original

length is 12

CM and a change in temperature final

minus initial 95us 15 is 80 so the

temperature increases by

80 so the change in length is going to

be

02784

CM so what is the new

L the new L is going to be l0 plus the

changing left so that's going to be 12

02784 now let's do the same thing for

the width let's calculate the change in

width so that's going to be Alpha which

is 29 * 10 10-

6 times instead of l0 it's going to be

w0 which is

8 and the change in temperature is still

positive

80 so this is equal to

[Music]

01856

cm so now what is the new

width so w which is W initial plus the

change is now

8.01

856 so now we can calculate the new area

which is simply the new length times the

new

WID so let's make some space

first so it's going to be the new length

which is

12.27

eight4 times the new width which is uh

8.01

856 it's not going to change much but it

does increase a

little so the new area is now

96.44 6 if we round it square meters

so therefore the change in area is the

difference between these two values so

Delta

a is about

446 square

meters if you want to come up with a

formula that can help you to get this

answer here's what you need to

do

the formula is Delta a is equal to l0

w0 * 2 Alpha Delta

t plus Alpha 2 Delta t^2 now because

Alpha is so small it's 29 * 10^ - 6

Alpha squar is going to be very small so

the contribution of this term is

negligible so we could say that Delta a

is approximately equal to l0

w0 time 2 Alpha Delta

T now l0 is

12 w0 is

8

Alpha is 29 * 10- 6

and since I'm running out of space let's

move this somewhere

else and the change in temperature is

80 so 2 * 12 * 8 * 29 * 10 - 6 *

80 is

about.

4454 as you can see this answer is very

close to this

one now keep in mind I rounded this

answer so they're very close so that's

how you can approximate Delta

a but if you want to derive the equation

here's what you need to

do so we know that Delta L is equal

to

Alpha l0 Delta C and the new L value

is l0 plus Delta L where Delta L is

Alpha l0 Delta T so we can factor out

l0 if we do it's going to be 1 + Alpha

Delta

T so that's L if that's L we could say

that W is W * 1 + Alpha Delta

t

now the original area is the original

length times the original width the new

area is the new L times the new wi so

therefore the change in area is the

difference

between the new area and the original

area which is LW minus L

wo

so Delta

a is equal

to the new L times the new W now the new

L is basically this

value it's

l0 1 + Alpha Delta

T so we replace L with that new

value now we're going to replace

W with what it's equal to here so

w is

wo

Time 1 + Alpha Delta

T and then minus l0 w0 so notice that we

can take out an l0 and we can Factor out

a

w0 so let's go ahead and do that so this

is going to be l0

w0 and notice that we have these two

terms which since we have two of them we

can write them as 1 + Alpha Delta T

squar since we have two of

them and we took out both l0 and w0 so

we're left

with1 so now let's uh clear away some

space

so now what we're going to do at this

point to

find the new Delta a let's foil this

term so 1 + Alpha * delta T * another 1

+ Alpha * delta

T it's going to be 1 * 1 which is

1 and 1 * Alpha Delta delta T that's

simply Alpha Delta T time plus Alpha

Delta T * 1 which is another Alpha Delta

T and finally Alpha Delta T * Alpha

Delta T that's Alpha 2 Delta t^

2 so we're going to

replace this term with this expression

now keep in mind we can add these two

terms so it's really one + 2 Alpha Delta

t + Alpha 2 delta T 2

so the change in area is equal to the

original length times the original width

and then 1 + 2 Alpha Delta t plus Alpha

2 Delta t^ 2 and now let's not forget

about this one that we have here so

minus

one now we can cancel these two and so

we have the

equation that we had

before

so that's uh l0

w0 to Alpha Delta t plus alpha s Delta

t^

2 so when we use the equation l0 w0 and

times 2 Alpha Delta T we got an answer

that was like.

4454 for the change in area let's see

what will happen if we this

term so it's going to be 12 *

8 and then 2 * 29 * 10 - 6 the change in

temperature was

80 plus 29 * 10 - 6^ 2 * 80

2 2 * 29 * 10 - 6 *

80 this part is about

0.464 29 * 10 - 6 * 80 and then if you

square it you're going to get a very

small number of

0. 0 0 5 38 2 and 12 * 8 is

96

so including this term this is going to

be equal to

44

59

6 which rounds to the first answer that

we have

446 so as you can see the contribution

from this term is relatively

insignificant so thus we have the

equation the change in

area is approximately equal to 2 l0 w0

Alpha Delta

T now what about volume expansion so

let's say if we have a rectangular

solid the volume of this solid is equal

to the length time the width time the

height now if you raise the temperature

the entire volume will increase the

length will increase the width will

increase and the height will increase it

turns out that the equation that you can

use is this equation Delta V the change

in volume is equal to the coefficient of

volume

expansion times the original volume

times the change in

temperature now for most solids B is

approximately 3 time

Alpha if you go to the textbook you

could see that it's roughly the

case it may not be precise but it's

approximately three times

Alpha for example

aluminum has an alpha value of 25 * 10-

6 but it's beta value is about 75 * 10-

6 anytime you raise the temperature the

volume will increase if you decrease the

temperature the volume will

decrease so let's work on the problem by

the way this applies for not only solids

but liquids as

well a rectangular solid of brass has a

coefficient of volume expansion of 56 *

10 - 6 and I forgot to put the units

that's Celsius to

theus1 so this is our beta

value now we have the dimensions of the

rectangle so we can calculate the

initial volume it's going to be the

length time the width time the

height so that's going to be 5 * 6 *

8 5 * 6 is 30 and 3 * 8 is 24 so 30 * 8

is

240 and it's going to be cubit

feet * * 3 * is Cub feet so that's the

initial

volume so now let's calculate the change

in volume using this equation so it's

going to

be beta the coefficient of volume

expansion times the original volume

times the change in

temperature so B is 56 * 10- 6 the

initial volume is 240 cubic feet and the

change in temperature final minus

initial 90 - 10 which is a change of

80 so the change in

volume is 1

1.75 cubic

feet so the new volume is the original

Volume Plus the change in volume so

that's

240 plus

1.75 so the new volume is

241.7 cubic

feet

here's another one a cup contains 85

Mill of water at 80° C what is the new

volume at 15 so we can see that the

temperature is

decreasing so the volume of water should

decrease now we have the coefficient of

volume expansion is 210 * 10- 6 so let's

use the equation the change in volume is

equal to the original volume times the

coefficient times the change in

temperature so the coefficient is 210 *

10-

6 it's pretty large relative to the

solids now the original volume is 85

Mill and the change in temperature the

final temperature is 15 the initial

temperature is 80 so the change is

-65 which means that the change in

volume will be

negative

so the change in volume is about

1.16 milliliters so the new volume which

is the original Volume Plus the change

that's going to be

85+

1.16 or 85 minus

1.16 so the new volume is

83.8 4 milliters

so as you can see it decreased

slightly now earlier we talked

about area and how to calculate it using

Alpha this time we're going to talk

about how to calculate the change in

volume using Alpha as

well the new volume is the length time

the width time the height the original

volume is the original length times the

original width times the original height

so therefore we could say that the

change in volume which is V minus V

that's

LW minus l o

w now we know that the new volume

L from earlier in this video we said

it's uh l0 1 + Alpha * delta

T now now W is going to be something

similar it's wo * 1 + Alpha Delta T and

H if we follow the same pattern it's

going to be ho 1 + Alpha Delta T minus l

w

h so we're going to take out the GCF

we're going to factor out L W

ho if we do that it's going to be these

three

which we can write it as 1 + Alpha Delta

t to the third

power and if we take out these three

variables we're going to be left over

with a

one so this is the change in volume

notice that we can

replace L W ho with v so the change in

volume is V

* 1 + Alpha Delta

T raised to the 3

power minus

one now let's foil this three times so

it's 1 + Alpha Delta T time another 1 +

Alpha Del

times another

one now we've already foiled these two

earlier in the video we said it was 1 +

2 Alpha Delta

t+ Alpha 2 Delta t^

2 so now let's distribute it

to 1+ Alpha that's C on the right so one

times these three terms it's going to be

the same thing on on the left so it's

going to be 1 + 2 Alpha Delta

t+ alpha s Delta t^2 and

then Alpha Delta T times these three

terms it's going to be 1 * Alpha Delta T

is just Alpha Delta T and then these two

multiplied it's going to be 2 Alpha 2

Delta t^ 2 and then finally these two

multiplied will

be Alpha Cub * Delta TB and now let's

add the negative

1 so we can cancel the one and the

negative

1 we can combine these two terms so

that's going to be 3 Alpha Delta T and

we can combine these

two which is going to be 3 Alpha 2 Delta

t^ 2

and finally we have this one

remaining which is Alpha Cub Delta

CB so we can replace this

expression with 3 Alpha Delta

t

plus 3 Alpha

2 Delta t^ 2

plus Alpha Cub delta T

Cub now when Alpha is small particularly

for

solids we were able to cancel Alpha

squ because Alpha squ will be even

smaller so this term is

insignificant so if Alpha squ is very

small Alpha Cube it's even smaller so we

can get rid of this so we could say that

Delta V is V * 3 Alpha * delta T and we

said that beta is approximately 3 *

Alpha so then we come up with this

equation Delta V is equal to V * beta *

delta T which is the equation that we

were using for volume expansion so you

can see how it's related to linear

expansion so that is it for this video

thanks for watching and have a great day