Linear Expansion of Solids, Volume Contraction of Liquids, Thermal Physics Problems
Summary
TLDRThis 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
🔥 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.
📏 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.
📐 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.
💧 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.
📦 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.
🌡️ 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.
Mindmap
Keywords
💡Thermal Expansion
💡Coefficient of Linear Expansion (Alpha)
💡Delta L (ΔL)
💡Temperature Change (ΔT)
💡Length Contraction
💡Coefficient of Volume Expansion (Beta)
💡Change in Area (ΔA)
💡Change in Volume (ΔV)
💡Original Length (L0)
💡New Length
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
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