Thermal expansion of square from temperature increase of 50K

[ChiralSuperfields]

Homework Statement

Pleases see below

Relevant Equations

Please see below

For this problem,

The solution is,

I understand their logic for their equation, but when I was trying to solve this problem, I came up with a different expression:

$\Delta A = \Delta L_x\Delta L_y$
$\Delta L_x =\Delta L_y = \Delta L$ since this is a square.
$\Delta A = \Delta L^2$
$\Delta A = \alpha^2L_i^2\Delta T^2$

I also don't understand how how $\Delta T = 50K$ since it should be converted to celsius since the coefficient of linear expansion is in has unit of inverse celsius.

I did that and got $\Delta T = -223.15 °C$ from $T_k = T_c + 273.15$

Many thanks!

 

[mjc123]

The Kelvin scale and the Celsius scale have different zero points, but they have the same scale factor. Thus a temperature value of 50°C corresponds to a temperature of 323.15 K, but a temperature difference of 50°C corresponds to a temperature difference of 50 K.

A = L², but ΔA is not (ΔL)² but Δ(L²)
i.e. A - A_i = L_i²(1+αΔT)² - L_i²
=L_i²(2αΔT + α²(ΔT)²)
≈L_i²2αΔT if αΔT is small.

 

[erobz]

Yes, as @mjc123 points out; take care to write your change in area as follows:

$$(A + \Delta A) - A = ( L+\Delta L )^2 - L^2 $$

 

[kuruman]

$\Delta A = \Delta L_x\Delta L_y$

$A=L_xL_y.$
The product rule of differentiation is
$dA=L_yd(L_x)+L_xd(L_y)$.
It doesn't change when $d\rightarrow \Delta.$

 

[kuruman]

I also don't understand how how $\Delta T = 50K$ since it should be converted to celsius since the coefficient of linear expansion is in has unit of inverse celsius.

I did that and got $\Delta T = -223.15 °C$ from $T_k = T_c + 273.15$

Many thanks!

If $T_K=T_C+273.15,$ it follows that
$\Delta(T_K)=\Delta(T_C)+\Delta(273.15~\text{K}$).
How does 273.15 K change as the temperature of the sheet changes?

Spoiler

It doesn't change because it is a constant. So $\Delta(T_K)=\Delta(T_C).$

 

[ChiralSuperfields]

The Kelvin scale and the Celsius scale have different zero points, but they have the same scale factor. Thus a temperature value of 50°C corresponds to a temperature of 323.15 K, but a temperature difference of 50°C corresponds to a temperature difference of 50 K.

A = L², but ΔA is not (ΔL)² but Δ(L²)
i.e. A - A_i = L_i²(1+αΔT)² - L_i²
=L_i²(2αΔT + α²(ΔT)²)
≈L_i²2αΔT if αΔT is small.

Thank you for your reply @mjc123!

Sorry, where did you get L_i²(1+αΔT)² - L_i² from?

Many thanks!

 

[ChiralSuperfields]

Yes, as @mjc123 points out; take care to write your change in area as follows:

$$(A + \Delta A) - A = ( L+\Delta L )^2 - L^2 $$

Thank you for your reply @erobz !

So for the LHS of your expression,
$A_f = A + \Delta A$ and $A_i = A$, correct?

Where $\Delta A ≠ A_f - A_i$? Sorry I am still getting use to the notation.

Many thanks!

 

[ChiralSuperfields]

$A=L_xL_y.$
The product rule of differentiation is
$dA=L_yd(L_x)+L_xd(L_y)$.
It doesn't change when $d\rightarrow \Delta.$

Thank you for your reply @kuruman!

Interesting that you use the product rule. Is there a reason for that? Also, why would the area not change as the quantities are no longer differential?

Many thanks!

 

[ChiralSuperfields]

If $T_K=T_C+273.15,$ it follows that
$\Delta(T_K)=\Delta(T_C)+\Delta(273.15~\text{K}$).
How does 273.15 K change as the temperature of the sheet changes?

Spoiler

It doesn't change because it is a constant. So $\Delta(T_K)=\Delta(T_C).$

Thank you for your reply @kuruman!

That helps to see why the change in temperatures are directly proportional to each other.

However, I am curious why are you allowed to add delta on each side. I have never seen the algebraic operation before (that I am aware of)

Many thanks!

 

[kuruman]

Thank you for your reply @kuruman!

That helps to see why the change in temperatures are directly proportional to each other.

However, I am curious why are you allowed to add delta on each side. I have never seen the algebraic operation before (that I am aware of)

Many thanks!

Do you understand what $\Delta A$ symbolizes where $A$ is some quantity?

 

[erobz]

Thank you for your reply @erobz !

So for the LHS of your expression,
$A_f = A + \Delta A$ and $A_i = A$, correct?

Correct

Where $\Delta A ≠ A_f - A_i$? Sorry I am still getting use to the notation.

No, that is exactly what it is. How you managed to learn Calculus without understanding this notation is somewhat perplexing…

$A + \Delta A = ( L + \Delta L)^2$

If you subtract $A$ ( the area before the change) from both sides you are left with the change in $A$, i.e. $( \Delta A)$.

Then ( after expanding the RHS ) if you divide both sides by $\Delta L$ and take the limit as $\Delta L \to 0$ you have the formal method of finding the derivative of $A$ w.r.t. $L$.

 

[haruspex]

Thank you for your reply @kuruman!

That helps to see why the change in temperatures are directly proportional to each other.

However, I am curious why are you allowed to add delta on each side. I have never seen the algebraic operation before (that I am aware of)

Many thanks!

It's easily proved.
$T_{K, initial}= T_{C, initial}+273.15$
$T_{K, final}= T_{C, final}+273.15$
$T_{K, final}=T_{K, initial}+\Delta T_{K}$
etc. You can finish it.

 

[ChiralSuperfields]

Do you understand what $\Delta A$ symbolizes where $A$ is some quantity?

Thank you for your reply @kuruman!

I think $\Delta A$ symbolizes the change in $A$ so $A_f - A_i$.

Many thanks!

 

[ChiralSuperfields]

Correct

No, that is exactly what it is. How you managed to learn Calculus without understanding this notation is somewhat perplexing…

$A + \Delta A = ( L + \Delta L)^2$

If you subtract $A$ ( the area before the change) from both sides you are left with the change in $A$, i.e. $( \Delta A)$.

Then ( after expanding the RHS ) if you divide both sides by $\Delta L$ and take the limit as $\Delta L \to 0$ you have the formal method of finding the derivative of $A$ w.r.t. $L$.

Thank you for your reply @erobz!

I am relearning calculus at the moment.

So since $A = A_f - A_i$, then $A + \Delta A - A = A + A_f - A_i - A = A_f - A_i = \Delta A$, correct?

Many thanks!

 

[ChiralSuperfields]

It's easily proved.
$T_{K, initial}= T_{C, initial}+273.15$
$T_{K, final}= T_{C, final}+273.15$
$T_{K, final}=T_{K, initial}+\Delta T_{K}$
etc. You can finish it.

Thank you for your reply @haruspex!

I can see how to prove it.

Did you get the third line $T_{K, final}=T_{K, initial}+\Delta T_{K}$ by inspection i.e
$T_{K, final}=T_{K, initial}+ T_{K, final} - T_{K, initial}$
$0 = 0$

Many thanks!

 

[erobz]

Thank you for your reply @erobz!

I am relearning calculus at the moment.

So since $A = A_f - A_i$, then $A + \Delta A - A = A + A_f - A_i - A = A_f - A_i = \Delta A$, correct?

Many thanks!

Yeah, but there is no need to go though all that. You think of the $\Delta A$ as taking a step away from the current value of $A$, which is the result of taking a step $\Delta L$ away from the current value of $L$.

 

[ChiralSuperfields]

Yeah, but there is no need to go though all that. You think of the $\Delta A$ as a step away from the current value of $A$.

Thank you for your reply @erobz!

In calculus, isn't $\Delta A$ always positive?

Many thanks!

 

[erobz]

Thank you for your reply @erobz!

In calculus, isn't $\Delta A$ always positive?

Many thanks!

The notation covers steps in either direction.

 

[ChiralSuperfields]

The notation covers steps in either direction.

Oh, thank you for your reply @erobz!

 

[erobz]

Oh, thank you for your reply @erobz!

It’s usually written without regard for what direction is the step. It could be a positive value change just as easily as it could be a negative value change. The math handles it no differently…so you usually just see $+$ signs for simplicity. If that is what you are asking about it “always being positive in calculus ”.

If you had in mind that the change is always positive…no.

 

[ChiralSuperfields]

It’s usually written without regard for what direction is the step. It could be a positive value change just as easily as it could be a negative value change. The math handles it no differently…so you usually just see $+$ signs for simplicity. If that is what you are asking about it “always being positive in calculus ”.

If you had in mind that the change is always positive…no.

Ah ok. Thank you for help @erobz !

 

[kuruman]

I think $\Delta A$ symbolizes the change in $A$ so $A_f - A_i$.

Right. It is the difference between the final and initial value of the quantity. Now if this quantity is the sum of two quantities, say $A=B+C$, then a change in $A$ can be due to a change in $B$ or a change in $C$ or a change in both. In the third possibility the change in $A$ would be the sum of the changes in $B$ and $C$.

A concrete example of this is mechanical energy conservation. Mechanical energy is the sum of kinetic and potential energy, $ME=KE+PE$. If a block is sliding down a frictionless ramp, both kinetic and potential energy change but their sum (the mechanical energy) does not. The statement $\Delta (ME)=0$ says that mechanical energy is conserved (it doesn't change.) This means that the sum $KE+PE$ also doesn't change although individually both the kinetic and potential energies change. This can only mean that the sum of the changes must be zero, $\Delta (KE)+\Delta (PE)=0.$ Note that if one of the changes is positive, the other has to be negative.

I hope that this example clarifies why $A=B+C$ implies $\Delta A=\Delta B+\Delta C.$

 

[ChiralSuperfields]

Right. It is the difference between the final and initial value of the quantity. Now if this quantity is the sum of two quantities, say $A=B+C$, then a change in $A$ can be due to a change in $B$ or a change in $C$ or a change in both. In the third possibility the change in $A$ would be the sum of the changes in $B$ and $C$.

A concrete example of this is mechanical energy conservation. Mechanical energy is the sum of kinetic and potential energy, $ME=KE+PE$. If a block is sliding down a frictionless ramp, both kinetic and potential energy change but their sum (the mechanical energy) does not. The statement $\Delta (ME)=0$ says that mechanical energy is conserved (it doesn't change.) This means that the sum $KE+PE$ also doesn't change although individually both the kinetic and potential energies change. This can only mean that the sum of the changes must be zero, $\Delta (KE)+\Delta (PE)=0.$ Note that if one of the changes is positive, the other has to be negative.

I hope that this example clarifies why $A=B+C$ implies $\Delta A=\Delta B+\Delta C.$

Thank you for your reply @kuruman!

I see now how $A = B+C$ implies $\Delta A=\Delta B+\Delta C$ from that mechanical energy conservation example. So is there really no algebraic operation that allows to apply $\Delta$ to each quantity in an equation, it really just about it logically making sense?

Many thanks!

 

[haruspex]

Thank you for your reply @haruspex!

I can see how to prove it.

Did you get the third line $T_{K, final}=T_{K, initial}+\Delta T_{K}$ by inspection i.e
$T_{K, final}=T_{K, initial}+ T_{K, final} - T_{K, initial}$
$0 = 0$

Many thanks!

No, it continues:

It's easily proved.
$T_{K, initial}= T_{C, initial}+273.15$
$T_{K, final}= T_{C, final}+273.15$
$T_{K, final}=T_{K, initial}+\Delta T_{K}$

$T_{C, final}=T_{C, initial}+\Delta T_{C}$
$\Delta T_{K}-\Delta T_{C}=(T_{K, final}-T_{K, initial})-(T_{C, final}-T_{C, initial})$
$=273.15-273.15=0$

 

[ChiralSuperfields]

No, it continues:

$T_{C, final}=T_{C, initial}+\Delta T_{C}$
$\Delta T_{K}-\Delta T_{C}=(T_{K, final}-T_{K, initial})-(T_{C, final}-T_{C, initial})$
$=273.15-273.15=0$

Oh ok. Thank you for your help @haruspex !

 

[kuruman]

I see now how $A = B+C$ implies $\Delta A=\Delta B+\Delta C$ from that mechanical energy conservation example.

I am not entirely convinced that you do. Both @haruspex and I have addressed your other question

I also don't understand how how ΔT=50K since it should be converted to celsius since the coefficient of linear expansion is in has unit of inverse celsius.

using different approaches. Can you show how one gets the same result, $\Delta T_K=\Delta T_C$ using each approach?
The @haruspex approach, starting from
$T_{K, initial}= T_{C, initial}+273.15$
$T_{K, final}= T_{C, final}+273.15$
$T_{K, final}=T_{K, initial}+\Delta T_{K}$

The @kuruman approach, starting from
$A=B+C \implies \Delta A=\Delta B+\Delta C$
$T_{K}= T_{C}+273.15.$

Everything that you might need has already been discussed.

 

[ChiralSuperfields]

I am not entirely convinced that you do. Both @haruspex and I have addressed your other question

using different approaches. Can you show how one gets the same result, $\Delta T_K=\Delta T_C$ using each approach?
The @haruspex approach, starting from
$T_{K, initial}= T_{C, initial}+273.15$
$T_{K, final}= T_{C, final}+273.15$
$T_{K, final}=T_{K, initial}+\Delta T_{K}$

The @kuruman approach, starting from
$A=B+C \implies \Delta A=\Delta B+\Delta C$
$T_{K}= T_{C}+273.15.$

Everything that you might need has already been discussed.

Thank you for your reply @kuruman!

Using the @haruspex approach I would solve for 273.15 for the top two equations then set them equal to each other

$T_{K, initial} - T_{C, initial} = T_{K, final} - T_{C, final}$
$T_{K, initial} - T_{K, final} = - T_{C, final} + T_{C, initial}$
$-(T_{K, finial} - T_{K, initial}) = -(T_{C, final} - T_{C, initial}A)$
$-\Delta T_k = -\Delta T_C$
$\Delta T_k = \Delta T_C$

Using the @kuruman approch,
$T_{K}= T_{C}+273.15$ implies $\Delta T_{K} = \Delta T_{C}+ \Delta 273.15$
And since 273.15 is a constant,
$\Delta T_{K} = \Delta T_{C}+ 273.15$
$\Delta T_{K} = \Delta T_{C}$ So change in kelvin is proportional to change in Celsius.

Many thanks!

 

[haruspex]

$T_{K}= T_{C}+273.15$ implies $\Delta T_{K} = \Delta T_{C}+ \Delta 273.15$
And since 273.15 is a constant,
$\Delta T_{K} = \Delta T_{C}+ 273.15$

No, since 273.15 is a constant, $\Delta 273.15=0$, so $\Delta T_{K} = \Delta T_{C}$.

$\Delta T_{K} = \Delta T_{C}$ So change in kelvin is proportional to change in Celsius.

Not quite. $\Delta T_{K} = \Delta T_{C}$ So change in kelvin is equal to change in Celsius.

 

[ChiralSuperfields]

Thank you for your reply @haruspex!

No, since 273.15 is a constant, $\Delta 273.15=0$, so $\Delta T_{K} = \Delta T_{C}$.

Whoops, thank you for mentioning my mistake!

Not quite. $\Delta T_{K} = \Delta T_{C}$ So change in kelvin is equal to change in Celsius.

True, that is better wording

Many thanks!