Is a moving lump of ice hotter than the same ice in stationary?

According to what I've learnt in high school physics

Temperature is the average kenetic energy of the particles within a substance
Internal energy is the sum of K.E. and P.E. of all particles within the object considered

In a waterfall, the water at the bottom is slightly hotter (i.e. a slightly higher temp.) than that at the top because the P.E. of water is converted into K.E. which then is converted into sound and internal energy of the water below the waterfall according to mcΔT

Question:
Assume we have 2 lumps of ice of 0[sup]o[/sup]C
One lump is placed on a frictionless desk of 0[sup]o[/sup]C, stationary
Another lump, also on the same frictionless desk moves to the west with uniform velocity 1 m/s

By K.E. = 1/2mv^2

K.E. 1st lump of ice = 0J
K.E. 2nd lump of ice = 0.0005J

Maybe I'm still having some misconceptions, but does the temperature of both lumps of ice the same? (Assume no heat loss to surroundings)

Question 2:
Assume we have a 3rd lump of ice (identical to 1st and 2nd) in projectile motion
Take g = 10m/s^2
vhorizontal = 1m/s
vvertical = 1m/s
Angle = 45 (deg)
height from table = 0m

Would the temperature of the ice varies along its path, or would it remains the same? (Assume no heat loss to surroundings)

I"m not really clear on the "average" in the definition of "temperature", is it correct to assume all particles within the lumps of ice all moving with the same velocity? (ignore the vibrations)

P.S. The reason of asking these questions is to check whether modifying http://www.nature.com/nphoton/journal/v2/n8/full/nphoton.2008.145.html in making some cold based particle weapons for sci fi is physically sound

 

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Will a stationary lump of ice melt faster if you run by it quickly? You must remember that movement is relative. I've never heard that water at the top of a waterfall is slightly warmer than that at the bottom. PE and KE cannot be internal to the object, and are therefore different from heat energy. If this were not true I could observe a comet possessing huge kinetic energy as it flies through space, and then chase after it; when I match its velocity what exactly have I done to change its internal energy?

 

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In case 1, the stationary cube would be colder than the moving one, since the moving one would pick up heat from friction between the cube and the surface. Assumng constant motion as well as a host of other things.

In case 2 the stationary cube would remain colder and the moving one pick up heat from friction also, but since gravity would make it move faster as it goes, and it would heat at an increasing speed, due to the square of the engery in KE.

Not enough is given to compare the temperature of the two moving pieces of ice.

 

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Energy converted into sound does not raise the temperature of the water.
Energy dissipated by friction (even water to water friction) does raise the temperature of the water.

Yes. The temperature of each piece of ice WITH RESPECT TO AN OBSERVER ON THE ICE is the same. (Part in caps is important.) Once it hits something, however, part of the kinetic energy is translated into heat.

Same as above.

Yes.

 

A lump of ice moving RELATIVE TO WHAT?

 

"Yes. The temperature of each piece of ice WITH RESPECT TO AN OBSERVER ON THE ICE is the same. (Part in caps is important.) Once it hits something, however, part of the kinetic energy is translated into heat."

How does this account for the friction at the base converting to head as the ice cube moves? I believe this would increase the overal temerature of the moving ice to any observer.

 

I wish to reconsider by previouis post in which I said that the speed up due to gravity of the ice cube would create more heat. Coulomb's Law of Friction states that the friction of a moving body does not change with its speed.

Been wrong before. Hope to be wrong again.

 

I would think you would get a warmer brick if you throw it due to air resistance and skin friction.

 

Actually, there is ample literature that contradicts your statement. Some physicists (Einstein, Planck and Tolman) have put forward very compelling arguments for the transformation \(T=\gamma T'\) while others (Moller, Ott) argued for \(T=T'/\gamma\). Excepting Landsberg, no one supports \(T=T'\). For a good synopsis, see this paper.

 

yes.

 

The original question was not relativistic but classical, and the case involves friction and air resistance.


A complete discription of the sliding block in each case, would also have to take the heat generated by the above using diffrent frames of refrence.

Comments?

 

Landsberg is alone in this belief. The problem stems from agreeing what is the correct starting point for finding the transformation. Landsberg simply found a starting point that is not conducive for finding a transformation between frames. As you can see from the many papers on the subject different people found different laws of thermodynamics that lead to contradictory (different) transformations. The subject is not settled yet.

 

This is a simple engineering problem, not worthy of a great deal of theorizing.

Friction generates heat. Heat increases temperature.


See: http://en.wikipedia.org/wiki/Friction

or any engineerin text on the subject.

 

johnpepper, I'm fairly certain the OP was asking about KE, not friction.

Also, the comment about a frictionless desk and the following quote should make it clear what the intent was

No "heat loss to surroundings" most likely meant no heat gain, either. The core of the question is whether or not PE and KE are internal to an object and whether that energy is expressed in temperature. I gave my 2 cents in the first response

IMO, KE and PE are contained in a system, not an object. The system includes the measuring body.

 

OK,

Thanks

 

The portion of this statement in bold is not accurate under all realistic conditions. Under any circumstances where evaporation may occur as the water falls, it will be cooled in the process. This is most evident in dry hot climates, but should be observable so long as the surrounding air temperature is greater than that of the water and the humidity is not so great so as to prevent effective evaporation.

Prior to the introduction of refrigeration, the mechanism involved was used to make ice. Ice houses utilizing this process were in operation, in SW USA as late as the early 1970's. Perhaps longer.

It maybe that the cooling from evaporation is tempered to some extent by an increase in kinetic energy. However, from experience water at the bottom is most certainly not warmer than that at the top. Though it is true that in a waterfall it might be difficult to notice the difference, as the water in the mountains of my youth could not be considered warm in either location.