Change in volume of sphere with change in temperature

  • #1
MatinSAR
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Homework Statement
Two spheres of same size are made of the same metal but one is hollow and the other is solid. They are heated to same temperature, then what will happen?
Relevant Equations
Thermal expansion.
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  • #2
...equal change in radius does not mean the same volume change.
Explain your rationale.

If two spheres, both with radius r1 double their radius to r2,
what can we say about comparing their volumes?
 
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  • #3
DaveC426913 said:
Explain your rationale.

If two spheres, both with radius r1 double their radius to r2,
what can we say about comparing their volumes?
##V_2=8V_1##
 
  • #4
MatinSAR said:
##V_2=8V_1##
That's not what the question is saying.

Hollow Sphere HS1 has radius r1.
Solid Sphere SS1 has radius r1.
i.e. HV1 = SV1

As you said: "the radius of the two spheres changes equally"

So, after both spheres are heated, r1 increases to r2.What can we say about the relationship between HV2 = SV2?
 
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  • #5
DaveC426913 said:
That's not what the question is saying.

Hollow Sphere HS1 has radius r1.
Solid Sphere SS1 has radius r1.
i.e. HV1 = SV1

As you said: "the radius of the two spheres changes equally"

So, after both spheres are heated, r1 increases to r2.What can we say about the relationship between HV2 = SV2?
We can say that they are equal.

But what do you mean by ##r_1##?
outer radius of hallow sphere?
 
  • #6
Gotta say, I am a little concerned about the grammar on that website.

...its orientation will do not affect it's expansion...
...we can say both sphere will expand same...
...because material used for making both of them are same......
 
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  • #7
MatinSAR said:
We can say that they are equal.
So you acknowledge the text's answer then?
An equal final radius of both spheres results in an equal final volume.

MatinSAR said:
But what do you mean by ##r_1##?
No, r1 is simply the initial radius of both spheres before expansion.
And r2 is the radius of both spheres after expansion.
 
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  • #8
DaveC426913 said:
So you acknowledge the text's answer then?
An equal final radius of both spheres results in an equal final volume
No beacuse :
2023_09_24 5_18 PM Office Lens.jpg
##r_2-r_1 ## is same for both spheres but ##V_2-V_1## is not.
 
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  • #9
There is no mention in the problem about inner and outer radii.

A solid sphere of radius 1 has a volume of 4/3π.
A hollow sphere of radius 1 has a volume of 4/3π.
(If you sunk them both in a bathtub, they would both displace 4/3π units of water.)
 
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  • #10
DaveC426913 said:
There is no mention in the problem about inner and outer radii.

A solid sphere of radius 1 has a volume of 4/3π.
A hollow sphere of radius 1 has a volume of 4/3π.
It said that they have same size.
I think It means that apparent volume of hallow sphere equals volume of solid sphere. But real volume of hallow sphere is smaller than volume of solid sphere.
 
  • #11
MatinSAR said:
It said that they have same size.
I think It means that apparent volume of hallow sphere equals volume of solid sphere. But real volume of hallow sphere is smaller than volume of solid sphere.
There is no mention of apparent volume versus real volume in the problem.

You are overthinking it and what it is trying to teach you.
 
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  • #12
DaveC426913 said:
There is no mention of apparent volume versus real volume in the problem.

You are overthinking it and what it is trying to teach you.
So... In your opinion the spheres have same volume change. Am i right?
 
  • #13
MatinSAR said:
So... In your opinion the spheres have same volume change. Am i right?
It's not my opinion; it is what the problem intends.

As you are interpreting the problem, you have reached one technically correct solution. But the problem is not trying to teach you about apparent versus actual volume.

Volume - in this problem - is being considered simply as 4/3πr3. i.e. how much water it would displace if sunk in a bathtub.Part of learning is understanding the nature and scope of what, exactly, you are being asked to solve.
 
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  • #14
DaveC426913 said:
It's not my opinion; it is what the problem intends.

As you are interpreting the problem, you have reached one technically correct solution. But the problem is not trying to teach you about apparent versus actual volume.

Volume - in this problem - is being considered simply as 4/3πr3. i.e. how much water it would displace if sunk in a bathtub.Part of learning is understanding the nature and scope of what, exactly, you are being asked to solve.
Thanks for your help.
 
  • #15
I agree with @DaveC426913 that what you were thinking about is not what the problem intended. However, what you were thinking about is also very interesting. Call it “volume of metal”.

You note that the change in volume of metal is not the same for the two spheres. However, the original volume of metal is not at all equal, so perhaps that isn’t too surprising. So maybe the question ought to be is the change in volume of metal proportional? Is ## \frac {\Delta V} V## the same in the two cases? You’ll pretty easily find that it is, and, in fact, that is a general principle for any shape.
 
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  • #16
Cutter Ketch said:
I agree with @DaveC426913 that what you were thinking about is not what the problem intended.
If we examine the question in this way, it will be too simple, that's why I thought that this is not the point.
These two spheres have the same initial volume and temperature change, so it is more than obvious that they have equal expansion.

Cutter Ketch said:
You note that the change in volume of metal is not the same for the two spheres. However, the original volume of metal is not at all equal, so perhaps that isn’t too surprising. So maybe the question ought to be is the change in volume of metal proportional? Is ## \frac {\Delta V} V## the same in the two cases? You’ll pretty easily find that it is, and, in fact, that is a general principle for any shape.
Great idea!
 
  • #17
MatinSAR said:
If we examine the question in this way, it will be too simple, that's why I thought that this is not the point.
Except you are given the answer. And the answer is the simple one.

Now you're saying you don't like the simplicity of the question, so you'll answer it wrong, for the fun of it?

MatinSAR said:
These two spheres have the same initial volume and temperature change, so it is more than obvious that they have equal expansion.
Then why not move on to more challenging questions?
 
  • #18
DaveC426913 said:
Except you are given the answer. And the answer is the simple one.

Now you're saying you don't like the simplicity of the question, so you'll answer it wrong, for the fun of it?
I did not say that. I doubted the answer to the question for several reasons. And that's why I asked here.
I know that ##\Delta V = \beta V_i \Delta T##
But still I am not sure that what is ##V_i##. Is it real or apparent volume?
DaveC426913 said:
Then why not move on to more challenging questions?
Because I passed this lesson a year ago and Now I have harder lessons to pass.
 
  • #19
MatinSAR said:
I did not say that. I doubted the answer to the question for several reasons. And that's why I asked here.
But .. you are told the answer they're looking for. They even show you their math.

It makes no sense to have both the answer and the rationale given to you by the article (and by us), and yet continue question if the article "really" meant what it is really telling you. But you do you.
 
  • #20
DaveC426913 said:
But .. you are told the answer they're looking for. They even show you their math.

It makes no sense to have both the answer and the rationale given to you by the article (and by us), and yet continue question if the article "really" meant what it is really telling you. But you do you.
I agree.
Thanks for your help and time.
 
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FAQ: Change in volume of sphere with change in temperature

How does temperature affect the volume of a sphere?

Temperature affects the volume of a sphere through thermal expansion. When the temperature of a sphere increases, the sphere's material expands, causing its volume to increase. Conversely, when the temperature decreases, the material contracts, and the volume decreases.

What is the formula for calculating the change in volume of a sphere due to temperature change?

The change in volume (ΔV) of a sphere due to temperature change can be calculated using the formula: ΔV = V₀ * β * ΔT, where V₀ is the initial volume, β is the coefficient of volumetric thermal expansion for the material, and ΔT is the change in temperature.

What is the coefficient of volumetric thermal expansion?

The coefficient of volumetric thermal expansion (β) is a material-specific constant that quantifies how much the volume of a material changes per degree change in temperature. It is typically expressed in units of 1/°C or 1/K.

Are the changes in volume due to temperature significant for all materials?

No, the significance of volume changes due to temperature varies depending on the material. Materials with a high coefficient of volumetric thermal expansion, such as gases and some metals, exhibit more significant volume changes with temperature variations. Conversely, materials with a low coefficient, such as ceramics and some polymers, exhibit less significant changes.

How can we minimize the volume change of a sphere due to temperature fluctuations?

To minimize volume change due to temperature fluctuations, one can use materials with a low coefficient of volumetric thermal expansion. Additionally, maintaining a stable temperature environment or using engineering designs that accommodate thermal expansion can help mitigate the effects.

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