How does thermal expansion affect the energy of elastic materials?

In summary, the energy conversion in an elastic matter being stretched or relaxed involves a transfer of energy between potential energy (stored in the molecules as the separation of their mean positions increases or decreases) and kinetic energy (converted from work done against friction or stored as the molecules' motion increases or decreases). Additionally, in thermal expansion, the increase in kinetic energy of the molecules leads to an increase in potential energy due to the non-symmetric energy vs. separation curve of bonded atoms.
  • #1
abcd8989
44
0
I am puzzled with the energy conversion (actually I think I am easily get stuck with PE, KE and work done) when an elastic matter is being stretched or relaxed.

First of all, regarding the elastic hysteresis of a rubber, in the Force-extension graph, the area when the rubber is loaded is larger than that when unloaded. My interpretation is as follows. Please inspect whether I have any wrong or seemingly wrong concepts and kindly correct them.

While the rubber is being loaded, we do work on the rubber, by applying equal but opposite forces at the two end of the rubber, minor part of which is converted to the KE of the molecules as a result of work done against friction between molecular chains, as well as sound energy, major part of which is converted to the PE gained by the molecules as a result of the increased in separation of their mean positions. However, when it is being unloaded, the stored PE is converted to the KE of the molecules and sound energy. We do work to extend the rubber, which is lost to the surroundings. However, the rubber relaxes using its already stored PE, The difference in area is equal to the heat lost to the surroundings?

Secondly, regarding a matter which obeys Hooke's Law when being stretched (before the elastic limit), is there any gain in KE of the molecules of the matter? Or there is simply gain in strain energy? As the loading straight line in the force extension graph superpose with the unloading straight line, it seems that there is no heat lost to the surroundings.

Moreover, I don't understand the explanation for thermalexpansion using a simple PE-molecular separaion curve. The main idea is as follows: At absolute zero, the molecules have no KE and thus the molucules' PE is Uo, which is the lowest value. At a higher temperature, the molucules have some energy, which is above the minimum value Uo". The mean position of molecules is greater than the equilibrium position. When the temperature further increase, the mean position further increase non-symmetrically. As far as I know, it is the KE of the molucules. However, why does the increase in KE lead to an increase in PE? Is it because the mean separation of the molecules increases, so that the work done required to bring the molecules from infinity to a certain point decreases?

Much obliged for patient reading!
 
Physics news on Phys.org
  • #2
abcd8989 said:
I am puzzled with the energy conversion (actually I think I am easily get stuck with PE, KE and work done) when an elastic matter is being stretched or relaxed.

First of all, regarding the elastic hysteresis of a rubber, in the Force-extension graph, the area when the rubber is loaded is larger than that when unloaded. My interpretation is as follows. Please inspect whether I have any wrong or seemingly wrong concepts and kindly correct them.

While the rubber is being loaded, we do work on the rubber, by applying equal but opposite forces at the two end of the rubber, minor part of which is converted to the KE of the molecules as a result of work done against friction between molecular chains, as well as sound energy, major part of which is converted to the PE gained by the molecules as a result of the increased in separation of their mean positions. However, when it is being unloaded, the stored PE is converted to the KE of the molecules and sound energy. We do work to extend the rubber, which is lost to the surroundings. However, the rubber relaxes using its already stored PE, The difference in area is equal to the heat lost to the surroundings?

Secondly, regarding a matter which obeys Hooke's Law when being stretched (before the elastic limit), is there any gain in KE of the molecules of the matter? Or there is simply gain in strain energy? As the loading straight line in the force extension graph superpose with the unloading straight line, it seems that there is no heat lost to the surroundings.

Moreover, I don't understand the explanation for thermalexpansion using a simple PE-molecular separaion curve. The main idea is as follows: At absolute zero, the molecules have no KE and thus the molucules' PE is Uo, which is the lowest value. At a higher temperature, the molucules have some energy, which is above the minimum value Uo". The mean position of molecules is greater than the equilibrium position. When the temperature further increase, the mean position further increase non-symmetrically. As far as I know, it is the KE of the molucules. However, why does the increase in KE lead to an increase in PE? Is it because the mean separation of the molecules increases, so that the work done required to bring the molecules from infinity to a certain point decreases?

Much obliged for patient reading!

Your first two descriptions look fine.

On the thermal expansion issue: as you describe, the energy vs. separation curve of bonded atoms is not symmetric, though it is approximately parabolic for small deviations around the equilibrium point. (This is why elastic solids have the same stiffness in tension and in compression.)

I'm not sure what you mean when you say "it is the KE of the molecules." Also, no work is necessary to bring the atoms to the equilibrium point; they are attracted there spontaneously. But other than that, your description looks fine and is in agreement with the general understanding of atomic spacing.
 

1. What is the definition of energy of elastic matters?

The energy of elastic matters is the potential energy stored in an object when it is deformed or stretched. It is the result of the work done to change the shape or position of the object, and it can be released when the object returns to its original shape or position.

2. How is the energy of elastic matters calculated?

The energy of elastic matters can be calculated using the formula E = ½kx², where E is the energy, k is the spring constant, and x is the displacement of the object from its equilibrium position. This formula is derived from Hooke's law, which states that the force required to stretch or compress an elastic object is directly proportional to the distance it is stretched or compressed.

3. What factors affect the energy of elastic matters?

The energy of elastic matters is affected by the material properties of the object, such as its elasticity and stiffness. It is also influenced by the amount of force applied to the object, as well as the distance it is stretched or compressed.

4. How is the energy of elastic matters used in everyday life?

The energy of elastic matters is used in many everyday objects and activities. For example, it is used in rubber bands, springs, and trampolines. It is also used in sports equipment, such as tennis rackets and golf clubs, to provide a greater force when hitting a ball. Additionally, the energy of elastic matters is utilized in medical devices, such as braces and prosthetics.

5. Can the energy of elastic matters be converted into other forms of energy?

Yes, the energy of elastic matters can be converted into other forms of energy, such as kinetic energy. When an elastic object is released from its stretched or compressed state, the stored energy is converted into motion, resulting in kinetic energy. This conversion of energy is seen in actions like shooting a rubber band or jumping on a trampoline.

Similar threads

Replies
9
Views
2K
Replies
16
Views
2K
Replies
5
Views
1K
  • Special and General Relativity
Replies
27
Views
315
Replies
5
Views
851
Replies
4
Views
2K
Replies
3
Views
2K
Replies
17
Views
2K
Replies
22
Views
2K
Replies
26
Views
6K
Back
Top