# Heat produced on shaking

1. Mar 7, 2005

### vinter

OK, here is a question I found in a book some time ago :-

An insulated box containing a monatomic gas of molar mass M moving with a velocity v is suddenly stopped. Find the increment in the gas temperature as a result of stopping the box.

I thought on this and what came to my mind was this - I should first know the the exact definition of temperature to tackle this question. As I know, temperature is defined very precisely by the zeroth law of thermodynamics. But here, I need a definition based on the energy of the molecules. Of course, temperature is directly proportional to the total internal energy of the gas. And what is internal energy? It is the sum of several forms of mechanical energies one of them being the kinetic energy. Now, we know that the KE depends on the frame of reference, it comes that the temperature should also depend on the frame of reference. But does it? This is one question that comes from the main question.

The problem is not as simple as it seems.

I asked several people the same question but all did some baseless energy conservation calculations and actually reached the answer. No one seems to be realising that there is something serious here.

If you reach the answer, then you may like to think on the validity of your arguments by applying the same method to a slightly modified question - instead of the container stopping to zero velocity, its velocity suddenly changes to -v. Now find the change in temperature.

Help please!

2. Mar 7, 2005

### Clausius2

An interesting question. Thermodynamic temperature doesn't depend on the reference frame. That temperature is based on the <mean> temperature of the molecules and is measured somehow by definition in the reference frame attached to the center of mass of the group of molecules taken into account. It has no sense asking oneself about the thermodynamic temperature viewed from another reference frame, because that temperature won't represent the real thermal state of the substance.

If the container is moving at velocity $$v$$ towards the right, it will have two components of energy:

-One measured in the reference frame I said above, and it represents the local thermodynamic state: $$U_1=mc_vT_1$$ is the internal energy which contains the microscopic kinetic energy of the molecules. Under the assumption of local thermodynamic equilibrium there exists a mean velocity of fluctuation of the molecules, which is given by the Kinetic Theory.

-Another measured in the reference frame you want, always if you are coherent with the rest of the calculation. I have chosen the reference frame attached to ground. It represents the macroscopic kinetic energy: $$E_1=1/2 m v^2$$

Once the container have been stopped, its total energy is only the thermal energy based on the local thermodynamic properties: $$U_2=mc_vT_2$$

Under the assumption of adiabatic walls, the container doesn't exhange any heat with the external environment. And suppossing the effects of internal (viscous dissipation) are negligible, the desacceleration process can be approximated as an isentropic one:

$$U_1+E_1=U_2$$

which gives us an important result:

$$T_1+\frac{v^2}{2c_v}=T_2$$

This can be reshaped into a similar form:

$$\frac{T_2}{T_1}=1+\frac{v^2}{2c_vT_1}$$

This last result is similar to one found in Gas Dynamics Theory. The last term is of the order of the Mach Number. If you study Gas Dynamics some time, you will realise that the behavior of a compressible fluid particle inside a flow when being accelerated/desaccelerated, can be described with the whole approximation of this container problem. The temperature $$T_2$$ represents the stagnation temperature of the flow, and it depends by this proper definition on the reference frame chosen. On the other hand, the temperature $$T_1$$ is the so called Static temperature, and it is only a function of the local thermodynamic state as I said.

I hope this helps you. Otherwise post your doubts.

3. Mar 7, 2005

### Timbuqtu

First of all temperature is NOT defined by the average kinetic energy of a gas. It is defined as the proportionality between change of entropy and change of energy.

But yes, in case of an ideal gas the temperature is proportional to the average kinetic energy of the gas. But in an ideal gas the particle-motion is isotropic, the same number of particles moving to the left as moving to the right. So only is the frame of reference in which the average velocity is zero the temperature is proportional to the average kinetic energy. So the answer is no: temperature doesn't depend on the frame of reference.

4. Mar 8, 2005

### vinter

Thanks Clausius for your reply. But does your equation apply to my second problem? Suppose the container, instead of stopping completely, changes its velocity to -v. In that case, your $$E_{1}$$ term does not change. Will the temperature not change in that case?

Secondly, I would prefer a much rigorous treatment taking care of the definition of temperature. First, what's the definition? Is it -
The avg. KE of the molecules as seen from the centre of mass reference frame?

If yes, then I would prefer to think like :-
The initial temperature = avg. of half * m * (v-v0)^2
where v0 is the velocity of the centre of mass from the ground frame
Similarly, find the expression for the final temperature and then see what types of relations they exhibit.

5. Mar 8, 2005

### vinter

So do you mean, that (in the classical sense) if you have been told the KE's of all the molecules of a gas, and their masses also, then you can't find the temperature?

6. Mar 8, 2005

### Clausius2

Of course not if you assume an isentropic acceleration/decceleration. What happens in real world?. Although you could suppose adiabatic walls, there will be internal irreversibilities such as viscous dissipation which enhances a greater temperature at the final of the process.

As TumbuctÃº has said, the definition of temperature is the partial of the internal energy respect to the entropy remaining volume constant. In some way it will have to match with the result of the kinetic theory in which the root mean square of the molecular velocity is proportional to the square root of the temperature. In other words, temperature is proportional to the square of the rms velocity and so to the average kinetic energy.

Leave me some time to see if I am able to explain this mathematically.

7. Jul 15, 2005

### toocool_sashi

This is a JEE 2 marks question hehe :D and yes i did conservation of energy without thinking too much. I always had this small niggling doubt that Kinetic Energy of a system depends on your frame of reference. I cannot give an alternate definition of temperature or anything of that sort but i can say that if the velocity changes from v to -v there will be no temperature change. Firstly let this be clear...we are talking abt temperature change of box. now we have decreased the velocity from v to 0 and the temperature of box can be calculated from conservation of energy(friction, sound dissipation absent). i do not know anything abt temperature's relevance to frame...i just know that if conservation of energy holds...i can calculate temperature using conservation principle. immediately heat transfers from the box to the surrounding to maintain thermal equilibrium. now when box is again given a velocity of -v ... it gains KE and logically...and of course mathematically...box shud attain same temp as it had when it had velocity of v.

Last edited: Jul 15, 2005
8. Jul 15, 2005

### Gokul43201

Staff Emeritus
You're forgetting that work needs to be done by some external body to bring the box to rest. How do you use energy conservation without including this term ?

Last edited: Jul 15, 2005
9. Jul 15, 2005

### toocool_sashi

Here conservation of energy implies the 7th class stuff that we learn, "Energy in an isolated system must remain conserved". Initially, only energy is KE of the box. Finally, only energy wud be the heat energy. we equate both of them.

10. Jul 15, 2005

### Gokul43201

Staff Emeritus
How is the system isolated if an external body is interacting with it ? And since when did JEE start asking "7th class stuff" ?

11. Jul 15, 2005

### vinter

Hey are you all Indians here like me? If not, how do you know about JEE. In case you want to tell that it's a very prestigious exam and is famous throughout the world then do tell me because I just qualified in the exam one month ago ;-)
And yes, applying energy conservation is not that simple here, toocoolsashi. Indeed, the system is isolated and you can use conservation of energy, i.e., the total energy = constant. Now, some part of this total energy is related to the temperature and some to miscelleneous stuff like KE.
So, you have,
energy related to temperature + miscelleneous stuff = constant.
To get anything out of this equation, you need to know too much about the first term. You need to be aware of the relation between temp. and energy, for example. And this is not simple, since you may say that temperature is just a proportionality constant times the avg. KE of the molecules, but, KE from which reference frame? Ground reference frame? Or the frame of the container? Or that of the Zem mattresses of the planet Squornshellous Beta?

12. Jul 15, 2005

### Gokul43201

Staff Emeritus
Not all. And the JEE is virtually unknown outside India.

Please explain to me how this system is isolated when there is an external agent exchanging momentum (and hence, energy) with the body. You can not bring something to rest without doing work on it.

Even jolly ol' Adams can tell you that temperature is defined (approximately, or for an ideal gas) as the average KE in the CoM frame of the gas (or in this case, the rest frame of the box). If it weren't defined this way, there would be all kinds of double-counting and make a mess of energy conservation.

It's not hard to calculate the work done by the external body - just use momentum conservation. Of course, this may lead to surprising results, but such is physics.

13. Jul 15, 2005

### Gokul43201

Staff Emeritus
I believe there is no increment in the temperature, but let me find some time to write up a proof.

Last edited: Jul 15, 2005
14. Jul 15, 2005

### krab

The word "suddenly" is very important here. Consider that the box is stopped in a time that is short compared with the time it takes a sound wave to travel the length of the box. Then stopping requires no work (except for the work on the mass of the box itself), and all the kinetic energy of the uniform motion of the gas is translated first into a sound wave, then gradually as the sound dissipates, into a temperature rise.

OTOH, if the deceleration takes a long time, the kinetic energy of the gas is allowed to do work on the decelerator, and in the limit, there is no temperature rise.

15. Jul 15, 2005

### Gokul43201

Staff Emeritus
Krab, I'm not sure I follow this argument entirely. Just after the box is stopped "instantaneously", the molecules of the gas will have a net momentum (=Nmv_0, for N molecules, each of mass m) and hence a sound wave propagates through the box. How is this net momentum erased without the addition of an external impulse ?

16. Jul 15, 2005

### krab

It's like throwing a ball of putty against a wall. The momentum is transferred to the earth. But the energy heats the putty.

17. Jul 15, 2005

### Gokul43201

Staff Emeritus
Does it really ? How do you explain this physically ? Why does the wall not get heated as well ? I can't see how you can transfer momentum without transferring energy.

18. Jul 15, 2005

### krab

Old question. Often answered.

https://www.physicsforums.com/showthread.php?t=74337
https://www.physicsforums.com/showthread.php?t=51321

Momentum transferred is $M\delta v=mv$ (M is mass of earth and m is mass of gas.) Energy transferred is $M(\delta v)^2/2$, which is utterly negligible compared with $mv^2/2$; it is in fact smaller by a factor m/M (do the math). So all the energy except about 1 part in $10^{26}$ stays locally.

19. Jul 15, 2005

### Gokul43201

Staff Emeritus
Only, not in either of those threads.

I completely agree. The KE transfered to the Earth is negligible (By the way, where do we have the earth in the OP's problem ? Could the box not be stopped by another box?)

In an elastic collision. The putty getting smashed into a wall is hardly elastic. What about the strain and heat energies ?

Anyway, I don't think this is pertinent to this problem because I don't see why the box must be stopped by an object that is rigidly coupled to the earth.

I guess I shall. I may end up proving myself wrong after all.

20. Jul 15, 2005

### Staff: Mentor

An elastic collision is hardly necessary for the energy to remain in the putty. The translational kinetic energy of the putty is transformed into internal energy. Since the point of application of the wall's force on the putty does not move, that force does no real work on the putty and thus does change its total energy (to the degree that krab estimated).

A similar situation involving deformable bodies is that of person vertically jumping off the ground. The ground exerts a force on the person, but again that force does no real work. In this case the internal (chemical) energy of the person is transformed into KE of the center of mass.

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