How Does Potential Energy Dissipation Affect Temperature in a Metal Cube?

[leo9999]

Earlier today I've attended a physics exam and there is a query I'm not sure about.

A metallic cube (specific heat capacity 30 cal/K*Kg ) falls from an height of 50 m on a non-conducting surface, and it stops. After the inelastic collision, what is the temperature of the cube?

a_ The temperature doesn't change because the surface is non-conducting.
b_ We cannot find the temperature because we don't know the cube mass.
c_ It raises ( +3°K).
d_ It decreases ( -5°K).

In my opinion the correct answer is B. In fact, potential energy has to be dissipated and since the surface is non-conducting, there won't be a heat flow from the cube to the surface. So all potential energy will be used to increase the cube internal energy, then the temperature has to raise.
Without the cube mass, we can't calculate how much the temperature increases, but it should increase.
Am I right?
Thank you and sorry for my english!

 

[kuruman]

You are tight in the sense that you have identified the answer they expect you to give. In my opinion this is a poorly written question. Answer (a) implies that no heat transfer takes place because the surface does not allow heat to cross its boundary. That does not properly follow the energy transformations. Here is why.
1. Potential energy is converted to kinetic energy.
2. Kinetic energy of the cube's center of mass goes to zero when the block stops. The original kinetic energy is distributed among the atoms in the block which vibrate more rapidly in an increased number of ways from before. This increase is interpreted macroscopically as an increase in temperature.
3. Although it is true that the non-conducting surface will not take up any heat by conduction from the block, it is also true that the temperature of the surface will increase because the collision will deform it and set the its atoms vibrating more rapidly.

The author of the question should have just said "The temperature does not change" in (a). As for (b), not knowing the cube's mass is only one reason why we cannot find the temperature change. Even if someone gave us the mass, so that we can get a number for the available energy $mgh$, we still wouldn't know how that energy will be divided between the block and the surface.

 

[Chestermiller]

The corrrect answer is C.

Applying the first law of thermodynamics to this system, we have: $$\Delta U+\Delta (PE)=0$$ (since no deformational work is done on the cube and no heat exchange takes place with the surroundings). So, from this it follows that: $$mC\Delta T+mg(h_f-h_i)=0$$The m's cancel.

 

[kuruman]

The m's cancel.

Oops.

 

[haruspex]

since no deformational work is done on the cube

Wouldn't most deformational work on the cube end up as heat anyway? It's only work that might take some other stored, and perhaps recoverable, form that would divert some of the work done. Or maybe "deformational work" is defined to mean only that.

The question ought to specify that no work is done on the impacted surface. It is not sufficient to state that it is nonconducting.

 

[Chestermiller]

Wouldn't most deformational work on the cube end up as heat anyway?

Eventually, the increase in internal energy (i.e., temperature increase) of the cube would be transferred to the surrounding air as heat, which would then be dispersed to the rest of the surroundings.

It's only work that might take some other stored, and perhaps recoverable, form that would divert some of the work done. Or maybe "deformational work" is defined to mean only that.

The problem statement implies that the inelastic collision results in all the potential energy converting to internal energy. This is how a thermodynamics book would have asked the question.

The question ought to specify that no work is done on the impacted surface. It is not sufficient to state that it is nonconducting.

The problem statement is definitely "problematic." It should definitely have said that the impacted surface is rigid.

 

[haruspex]

Eventually, the increase in internal energy (i.e., temperature increase) of the cube would be transferred to the surrounding air as heat, which would then be dispersed to the rest of the surroundings.

Yes, sorry - sloppy wording on my part.
I meant, Wouldn't most deformational work on the cube end up as temperature rise anyway?

 

[Chestermiller]

I'd like to be a little more precise about this. As the cube is falling, it gains kinetic energy and loses potential energy, so its internal energy remains constant. Since this is an inelastic collision for the object, hitting the ground causes the kinetic energy to convert into deformational work on the object. The deformational work results in no elastic stored energy, and is all mechanically dissipated into internal energy.

 

[kuruman]

Since this is an inelastic collision for the object, hitting the ground causes the kinetic energy to convert into deformational work on the object.

What about the object with which the falling cube collides? Doesn't that dissipate a portion of the initial gravitational potential energy through the same mechanism? It takes two to effect a perfectly inelastic collision. The fact that the second object is a poor conductor means that if the two objects are at different temperatures, no heat is or will be exchanged between them. It does not mean that only one of the objects deforms mechanically during the collision.

 

[Chestermiller]

I think the surface is supposed to be regarded as rigid.