Cylinder & Mineral: Thermal Dilation

[Const@ntine]

Homework Statement

A hollow aluminum cylinder with a depth of 20.0 cm has an internal capacity of 2.000 L at 20.0 C. It's full with mineral turpentine, at 20.0 C. The two of them are heated slowly, until the temperature reaches 80.0 C.

a) How much of the mineral is spilled outside the cylinder?
b) How much mineral remains inside?
c) If the cylinder with the mineral from (b) is then again frozen to 20.0 C, how much wouldthe mineral be bellow the top of the cylinder?

Homework Equations

ΔL = αL_iΔT
ΔV = βV_iΔT
β = 3α

Aluminum: a = 4*10^-6 C^-1
Mineral: a = 9.0*10^-4 C^-1

The Attempt at a Solution

At first I figured I'd take the second formula, with Vi(mineral) = Vi(cylinder) = 2 L, find the Vf in both cases and then find the difference. But, that's not it. I did think of treating the depth as the "L" and going with the first formula to find out how "taller" each substance got, but that didn't work either. So... any ideas?

Any help is appreciated!

 

[rude man]

Homework Statement

A hollow aluminum cylinder with a depth of 20.0 cm has an internal capacity of 2.000 L at 20.0 C. It's full with mineral turpentine, at 20.0 C. The two of them are heated slowly, until the temperature reaches 80.0 C.

a) How much of the mineral is spilled outside the cylinder?
b) How much mineral remains inside?
c) If the cylinder with the mineral from (b) is then again frozen to 20.0 C, how much wouldthe mineral be bellow the top of the cylinder?

Homework Equations

ΔL = αL_iΔT
ΔV = βV_iΔT
β = 3α

Aluminum: a = 4*10^-6 C^-1
Mineral: a = 9.0*10^-4 C^-1

The Attempt at a Solution

At first I figured I'd take the second formula, with Vi(mineral) = Vi(cylinder) = 2 L, find the Vf in both cases and then find the difference. But, that's not it. I did think of treating the depth as the "L" and going with the first formula to find out how "taller" each substance got, but that didn't work either. So... any ideas?

Both the height and the diameter of the cylinder expand!

 

[Const@ntine]

Both the height and the diameter of the cylinder expand!

So I got done with (a) & (b), but I'm having a bit of trouble with (c).

Mineral: β = 9*10^-4 C^-1
Aluminum: a = 24*10^-6 C^-1

a)

Mineral: ΔV = βViΔT <=> ... <=> Vf = 2.108 * 10^-3 m³

Cylinder: ΔV = 3aViΔT <=> ... <=> Vf = 2.00864 * 10^-3 m³

ΔVf = 0.0996 * 10-3 m3 => 99.4 mL of mineral are spilled out. (same as the book's answer)

b) The cylinder has a new volume of Vf = 2.00864 * 10^-3 m³, so the mineral that is still left in the cylinder is 2.01 (same as the book's answer)

c) See, this is where I got stuck. First things first, the volumes change again (the new Vi are the Vf s from (a) ):

Mineral: ΔV = βViΔT <=> ... <=> Vf ' = 1.90146 * 10^-3 m³

Cylinder: ΔV = 3aViΔT <=> ... <=> Vf '= 2.001317 * 10^-3 m³

Thing is, I have to connect the volume with the height. The only formula I know that can do that is V = A*H. Now, I know only the H of the cylinder (Hi = 0.2 m). Since I've both heated him and frozen him, logically the height should change two times. Expand first, then shorten. So:

Cylinder:

ΔH = aHiΔT <=> ... <=> Hf = 0.200288 m

ΔH = aHiΔT <=> ... <=> Hf '= 0.199999 m

So I know that after everything, the cylinder's new height is Hf '. Problem is, I don't know anything about the mineral. I could use some help on how to approach this one. The book's answer is "0.998 cm".

 

[rude man]

So I got done with (a) & (b), but I'm having a bit of trouble with (c).
Mineral: β = 9*10^-4 C^-1
Aluminum: a = 24*10^-6 C^-1
a)
Mineral: ΔV = βViΔT <=> ... <=> Vf = 2.108 * 10^-3 m³
Cylinder: ΔV = 3aViΔT <=> ... <=> Vf = 2.00864 * 10^-3 m³
ΔVf = 0.0996 * 10-3 m3 => 99.4 mL of mineral are spilled out. (same as the book's answer)

How did you get 0.0996? It should be 0.9936 = 0.994. With small numbers like these the difference matters.

b) The cylinder has a new volume of Vf = 2.00864 * 10^-3 m³, so the mineral that is still left in the cylinder is 2.01 (same as the book's answer)

OK, rounded off, which is poor practice though.

c) See, this is where I got stuck. First things first, the volumes change again (the new Vi are the Vf s from (a) ):
Mineral: ΔV = βViΔT <=> ... <=> Vf ' = 1.90146 * 10^-3 m³

OK
You're rounding off too much all over, even if the book allows it.

Cylinder: ΔV = 3aViΔT <=> ... <=> Vf '= 2.001317 * 10^-3 m³

You don't need to recompute the volume of the cylinder at 20C, do you. It's still 2L.
So you wind up with 1.90146L fluid in a 2L container. And you can't figure out how much the liquid level is below the rim?

 

[Chestermiller]

At 20 C, what's the cross sectional area if the cylinder? In the final state, what is the cross sectional area of the mineral?

 

[Const@ntine]

How did you get 0.0996? It should be 0.9936 = 0.994. With small numbers like these the difference matters.

Mistake on my part, I meant to write 0.994.

OK, rounded off, which is poor practice though. OK
You're rounding off too much all over, even if the book allows it.

I've found that in cases where I don't follow the book's "rules" I get different results, which result in me doing the same exercise over and over trying to spot the mistake, when in reality it was just him taking less Significant Digits/rounding it off. So I just go with his version, just to be on the clear.

You don't need to recompute the volume of the cylinder at 20C, do you. It's still 2L.

Since I decrease the temperature, doesn't it shrink a bit, like how when I increased the temp it expanded?

So you wind up with 1.90146L fluid in a 2L container. And you can't figure out how much the liquid level is below the rim?

Kinda, yeah. I'm probably missing something, but I can't figure out what to do here. I know that at the very beginning, before the temperature modifications, the depth of the cylinder was 0.2 m. I can connect volume to height by V = A*H, but I did that* and I didn't get a correct result.

*I know that ΔA = 2aAiΔT. So I found Ai from: Vi = Ai*H <=> 2.000*10^-3 m³ = Ai * 0.2 m <=> Ai = 0.01 m²
Then I applied the formula to the two temperature modifications (from 20 C to 80 C, and from 80 C to 20 C), and by using V=A*H for the final numbers I tried to find the "height" of the mineral. But it didn't work.

At 20 C, what's the cross sectional area if the cylinder? In the final state, what is the cross sectional area of the mineral?

Well, at 20 C, the cross sectional area of the cylinder would be: Vi = Ai*H <=> 2.000*10^-3 m³ = Ai * 0.2 m <=> Ai = 0.01 m²

In the final state, I'm not sure how I can find the CSA of the mineral. I can only study the volume of it with my data. I'm probably missing something.

 

[Chestermiller]

Mistake on my part, I meant to write 0.994.
I've found that in cases where I don't follow the book's "rules" I get different results, which result in me doing the same exercise over and over trying to spot the mistake, when in reality it was just him taking less Significant Digits/rounding it off. So I just go with his version, just to be on the clear. Since I decrease the temperature, doesn't it shrink a bit, like how when I increased the temp it expanded?
Kinda, yeah. I'm probably missing something, but I can't figure out what to do here. I know that at the very beginning, before the temperature modifications, the depth of the cylinder was 0.2 m. I can connect volume to height by V = A*H, but I did that* and I didn't get a correct result.

*I know that ΔA = 2aAiΔT. So I found Ai from: Vi = Ai*H <=> 2.000*10^-3 m³ = Ai * 0.2 m <=> Ai = 0.01 m²
Then I applied the formula to the two temperature modifications (from 20 C to 80 C, and from 80 C to 20 C), and by using V=A*H for the final numbers I tried to find the "height" of the mineral. But it didn't work.
Well, at 20 C, the cross sectional area of the cylinder would be: Vi = Ai*H <=> 2.000*10^-3 m³ = Ai * 0.2 m <=> Ai = 0.01 m²

In the final state, I'm not sure how I can find the CSA of the mineral. I can only study the volume of it with my data. I'm probably missing something.

In the final state, the cross sectional area of the mineral has to be the same as that of the container. After all, it settles at the bottom of the container.

 

[Const@ntine]

In the final state, the cross sectional area of the mineral has to be the same as that of the container. After all, it settles at the bottom of the container.

Oh yeah, silly me. In that case:

Going by the Cylinder:

For T = 20 C to 80 C: ΔA = 2aAiΔT <=> Af = 0.0100288 m²

Now for T = 80 C to 20 C, where "Ai" = Af = 0.0100288 m²: ΔA = 2aAiΔT <=> Af '= 9.9712 * 10^-3 m²

So, the Cross Sectional Area of the mineral, at the end, is effectivelly Af ' = 9.9712 * 10^-3 m².

Now, I can take the problem in this direction:

>As shown above, the new height of the cylinder, after all the changes, is Hf '= 0.199999 m

>For the mineral, I have Vf ' = 1.910146 * 10^-3 m³ & Af ' = 9.9712 * 10^-3 m².

>I have the formula V = A*H

>So, for the mineral, I have: Vf ' = Af ' * Hm <=> Hm = 0.1906952022 m (m is for mineral).

>The Cylinder's final height is Hf ' = 0.199999 m

>So: DH = Hf' - Hm = 9.3037978 * 10-3 m = 0.930 cm

The book's answer is 0.998 cm. I guess the problem could be the rounding off of digits, but I went with the book's "rules", since its results were always rounded to 3 Significant Digits.

 

[rude man]

Since I decrease the temperature, doesn't it shrink a bit, like how when I increased the temp it expanded?

Yes, it shrinks back to what it was to begin with!

Kinda, yeah. I'm probably missing something, but I can't figure out what to do here. I know that at the very beginning, before the temperature modifications, the depth of the cylinder was 0.2 m. I can connect volume to height by V = A*H, but I did that* and I didn't get a correct result.

I did that and I got the correct result!
You know the new volume of the mineral oil (about 1.9L) and you know the volume of the cylinder (2L) and you don't need to know the area of the cylinder bottom since the walls are straight - please re-think this, this is AT MOST 9th grade algebra! (OK, extra hint: V = AH as you say, A is constant, so what is ΔH/H?)

 

[Const@ntine]

Yes, it shrinks back to what it was to begin with!

Yeah, after all the computations the end result is 0.1999999... m, so technically yes.

I did that and I got the correct result!
You know the new volume of the mineral oil (about 1.9L) and you know the volume of the cylinder (2L) and you don't need to know the area of the cylinder bottom since the walls are straight - please re-think this, this is AT MOST 9th grade algebra! (OK, extra hint: V = AH as you say, A is constant, so what is ΔH/H?)

Sorry, still not getting it to work. Best I can come up with is 0.931 cm, instead of 0.998 cm.

>We have the height of the Cylinder, H, at 0.2 m. We have its volume, V, at roughly 2.00 L = 2.00 * 10^-3 m³. We have its CSA, A, at 0.01 m².

>As for the mineral, we have the CSA as the Cylinder, a volume of 1.9 L = 1.9 * 10^-3 m³, and the unknown height.

Mineral: V = A*h <=> h = 1.9*10^-3 m³ / 0.01 m² = 0.19 m

So, ΔΗ = Η - h = 0.2 m - 0.19 m = 0.01 m = 1.0 cm

It's close enough to the book's 0.998 cm, but my problem is that even if I use my more detailed answers, I still can't get the "0.998 cm".

 

[Chestermiller]

The final cross sectional area at 20 C is the same as when it started at 20 C, 0.01

 

[rude man]

Mineral: V = A*h <=> h = 1.9*10^-3 m³ / 0.01 m² = 0.19 m0
So, ΔΗ = Η - h = 0.2 m - 0.19 m = 0.01 m = 1.0 cm
It's close enough to the book's 0.998 cm, but my problem is that even if I use my more detailed answers, I still can't get the "0.998 cm".

Your number for the mineral oil final volume V_f = 1.9L is not accurate, and that is because you have not carried out precision to the necessary extent, as I have urged several times. The exact number is V_f = 1.9002L so (2.0000 - 1.9002)/2.0000 x 20 cm = 0.998 cm.
I do think the book misled you into sticking to 3 decimal points. To get 0.998 you need to carry out your computations to 4.

 

[Const@ntine]

Your number for the mineral oil final volume V_f = 1.9L is not accurate, and that is because you have not carried out precision to the necessary extent, as I have urged several times. The exact number is V_f = 1.9002L so (2.0000 - 1.9002)/2.0000 x 20 cm = 0.998 cm.
I do think the book misled you into sticking to 3 decimal points. To get 0.998 you need to carry out your computations to 4.

Yeah, decimals are a bit of a sore spot. I know the rules behind the Significant Digits, but the book plays fast and loose with them, and I try to keep up with it.

Either way, thanks for the help everyone!