Doubts regarding conversion of mechanical work to heat

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I have problems understanding WHEN mechanical work will be converted to heat .

If a body moves , will there always be a rise in temperature , even if negligible ?

If not , then read this Problem I recently came across , with the solution too :

" From what height must a mass of ice be dropped so that the heat generated completely melts the ice block ? "

Now , if the ice is not encountering any obstruction . Does that mean , if the ice was dropped from a height above the minimum height , the ice would have melted some distance above , WITHOUT ENCOUNTERING AN OBSTRUCTION ?

Now if that is true , it suggests all bodies when set to motion have a rise in temperature .This contradicts my common sense.

Please enlighten.
 

Answers and Replies

  • #2
mathman
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Is the ice being dropped in a vacuum (shouldn't melt) or in air (melts due to air friction)?
 
  • #3
K^2
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No, a body in motion does not heat up. Friction, deformation, and compression (I might be forgetting something) convert mechanical energy into heat.

The way I understand the problem you stated, the ice cube is assumed to melt on impact. There is a minimum height from which it needs to fall to gain sufficient kinetic energy, so that impact with ground would cause the block to melt. Assuming, of course, that all heat energy goes into ice cube, and it is evenly distributed before the block shatters.
 
  • #4
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From the ice cube's point of view, energy would be appearing out of nothingness. Then again, I'm asking myself the following question: does conservation of energy only apply to inertial reference frames?
 
  • #5
fluidistic
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does conservation of energy only apply to inertial reference frames?
No. To isolated systems.
 
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  • #6
K^2
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mr. vodka, in non-inertial frames there are still conservation laws, but they have to do with Killing Vectors.

fluidstic, no, in general, energy is not conserved in accelerated reference frames. Though, if you found such a situation, it does point to problems in your choice of coordinate system.
 
  • #7
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Hm Killing Vectors, is new to me, daunting name, but it looks tasty on wikipedia, can't wait when we get to the serious classical mechanics in college :) Thanks

EDIT: in retrospection it should've been way more obvious that there's no law of conservation of energy in non-inertial frames (for people still unconvinced: imagine yourself in a universe with only one object and you're suddenly accelerated: the ball gains kinetic energy with nothing to compensate -- it seems like the essence of the law failing is the 3rd law of Newton failing)
 
  • #8
fluidistic
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fluidstic, no, in general, energy is not conserved in accelerated reference frames. Though, if you found such a situation, it does point to problems in your choice of coordinate system.

I'd like to see an example of an isolated system where energy is not conserved. I can't think of a single one at the moment. Can you give an example?
 
  • #9
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fluidistic, let me try to make a "realistic" example: isolate yourself and the earth from the rest of the universe. Jump out of a hot air balloon high in the sky: from your point of view the earth is gathering speed toward you, thus gaining kinetic energy. Is anything giving up energy in return?
 
  • #10
K^2
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I'd like to see an example of an isolated system where energy is not conserved. I can't think of a single one at the moment. Can you give an example?
Take a universe with a single object in it of mass m. Find inertial reference frame. Relative to that frame take a frame that accelerates at rate a. The energy of the object at any time t is (1/2)ma²t². Which also happens to be the total energy in that system. It is clearly not time-independent. Ergo, energy is not conserved.
 
  • #11
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Heat transferred through conduction and convection (i.e. all forms of heat except radiation) is itself mechanical work done by particles in motion. In a gas, heat is average KE of the moving particles due to collisions. In a solid, I believe the particles vibrate within a relatively fixed position, but the vibration is nonetheless motion and therefore performs work (work = force over a distance, correct?). I think if you want to analyze how any mechanical work at the macro level generates heat at the molecular level, you would just have to trace the energy transfers between the parts of the mechanical system at the level of direct contact between surfaces and the molecules that compose them.
 
  • #12
fluidistic
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fluidistic, let me try to make a "realistic" example: isolate yourself and the earth from the rest of the universe. Jump out of a hot air balloon high in the sky: from your point of view the earth is gathering speed toward you, thus gaining kinetic energy. Is anything giving up energy in return?
I'm not understanding well the situation. Am I over the surface of the Earth? If so, how does the Earth gather speed toward me? I understand that the center of mass of the system doesn't move so if the ballon moves, so does the Earth. The energy requiring for this motion comes from the molecules of air surrounding the balloon (they make it go higher and higher until the balloon oscillates around an equilibrium point).


Take a universe with a single object in it of mass m. Find inertial reference frame. Relative to that frame take a frame that accelerates at rate a. The energy of the object at any time t is (1/2)ma²t². Which also happens to be the total energy in that system. It is clearly not time-independent. Ergo, energy is not conserved.
I'm not sure it makes sense to choose a frame that "accelerates at rate a" from the object if there's nothing else in the universe. Its motion would be relative to... you? Then there's more than a single object. If I take an object of mass m as whole universe, I think it's meaningless to talk about any motion from it.
According to wikipedia
In the natural sciences an isolated system, as contrasted with an open system, is a physical system that does not interact with its surroundings. It obeys a number of conservation laws: its total energy and mass stay constant. They cannot enter or exit, but can only move around inside. An example is in the study of spacetime, where it is assumed that asymptotically flat spacetimes exist.
.
Now maybe I'm wrong of course and it might possible to choose a reference frame that is accelerating with respect to a single object universe. I'm just not sure.
 
  • #13
K^2
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I'm not sure it makes sense to choose a frame that "accelerates at rate a" from the object if there's nothing else in the universe. Its motion would be relative to... you? Then there's more than a single object. If I take an object of mass m as whole universe, I think it's meaningless to talk about any motion from it.
As I said, you first find an inertial frame. That is, any frame that moves at constant velocity relative to the object in question. That you can actually measure, can you not? Now choose ANY frame of reference that accelerates to the first frame of reference. In this new frame of reference, kinetic energy of the object is NOT constant.

All there is to it.

Oh, and person who write the article is either unfamiliar with General Relativity and Killing Vectors, or simply chose not to get into these details.
 
  • #14
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Sorry, my first example was ill (not wrong, but very unclear). Let me make it exact:

Imagine a universe with nothing but you and a ball with the same weight as you. Call the inertial frame of reference at the center of mass A, and call your non-inertial frame of reference B. It is clear that B is non-inertial, because you (and thus your frame) is accelerating relative to the inertial frame A.

[tex]U_i = -\frac{Gm^2}{d}[/tex] with d the initial distance between you and the ball

Now at the instant you and the ball are d/2 meter away from each other, call your kinetic energy (relative to the center of mass A) K. Of course the ball with the same mass has the same kinetic energy. Law of conservation of energy:

[tex]U_i = -\frac{Gm^2}{d} = -\frac{Gm^2}{d/2} + 2K[/tex]

Solving this for K gives [tex]K = \frac{Gm^2}{2d}[/tex] and since [tex]K = \frac{mv^2}{2}[/tex] this gives that [tex]v_{ball-rel.-to-A} = \sqrt{ \frac{Gm}{d} } = v_{you-rel.-to-A}[/tex].

Now relative to B (= your view), we use Galileo's addition of velocities: [tex]v_{ball-rel.-to-B} = v_{ball-rel.-to-A} + v_{you-rel.-to-A} = 2\sqrt{ \frac{Gm}{d} }[/tex]

Since in B the only thing moving is the ball, the total energy of the system according to B is
[tex]E_B = K_B + U_B = \frac{m v_{ball-rel.-to-B}^2}{2} - \frac{Gm^2}{d/2} = \frac{2Gm^2}{d} - \frac{Gm^2}{d/2}[/tex]

And this is NOT equal to the potential energy the system started with. Tadum :)
 
  • #15
fluidistic
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As I said, you first find an inertial frame. That is, any frame that moves at constant velocity relative to the object in question. That you can actually measure, can you not? Now choose ANY frame of reference that accelerates to the first frame of reference. In this new frame of reference, kinetic energy of the object is NOT constant.

All there is to it.

Oh, and person who write the article is either unfamiliar with General Relativity and Killing Vectors, or simply chose not to get into these details.
Ok now I understand perfectly, this is kind of troublesome for me.
Sorry, my first example was ill (not wrong, but very unclear). Let me make it exact:

Imagine a universe with nothing but you and a ball with the same weight as you. Call the inertial frame of reference at the center of mass A, and call your non-inertial frame of reference B. It is clear that B is non-inertial, because you (and thus your frame) is accelerating relative to the inertial frame A.

[tex]U_i = -\frac{Gm^2}{d}[/tex] with d the initial distance between you and the ball

Now at the instant you and the ball are d/2 meter away from each other, call your kinetic energy (relative to the center of mass A) K. Of course the ball with the same mass has the same kinetic energy. Law of conservation of energy:

[tex]U_i = -\frac{Gm^2}{d} = -\frac{Gm^2}{d/2} + 2K[/tex]

Solving this for K gives [tex]K = \frac{Gm^2}{2d}[/tex] and since [tex]K = \frac{mv^2}{2}[/tex] this gives that [tex]v_{ball-rel.-to-A} = \sqrt{ \frac{Gm}{d} } = v_{you-rel.-to-A}[/tex].

Now relative to B (= your view), we use Galileo's addition of velocities: [tex]v_{ball-rel.-to-B} = v_{ball-rel.-to-A} + v_{you-rel.-to-A} = 2\sqrt{ \frac{Gm}{d} }[/tex]

Since in B the only thing moving is the ball, the total energy of the system according to B is
[tex]E_B = K_B + U_B = \frac{m v_{ball-rel.-to-B}^2}{2} - \frac{Gm^2}{d/2} = \frac{2Gm^2}{d} - \frac{Gm^2}{d/2}[/tex]

And this is NOT equal to the potential energy the system started with. Tadum :)

Wow, nice example. Yeah I'm clueless as of now.
 
  • #16
K^2
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You can try Wikipedia's article on Killing Vectors, but it's all written in heavy GR jargon. Might be indistinguishable from Gaelic, but maybe worth a try.
 
  • #17
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But modern definition of temperature , even Wikipedia says ' temperature is a measure of the average kinetic energy of the molecules ' ....

Hence , when a body moves , it's total K.E increases . Hence average K.E of molecules also increases and thus temperature rises ?
 
  • #18
K^2
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No, it's kinetic energy in body's center of mass frame. So for each particle, it's contribution to temperature is going to be given by (1/2)*m*(v-vCM)², where vCM is the center of mass velocity.

Also, it's not a good definition of temperature. The correct definition involves energy distributions, rather than actual energies. But that doesn't really matter.
 
  • #19
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Then the temperature of the body does rise , even if very little , when set into motion ?
 
  • #20
K^2
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No. The average velocity relative to center of mass is what sets the temperature. If you increase velocity of all particles by the same amount, velocity relative to CM never changes.
 
  • #21
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Then the temperature of the body does rise , even if very little , when set into motion ?

I think it does as a result of force transfer between the point where force is being applied and all the other particles in the object. A cue ball hitting another ball must impart its energy into the target at a certain point of contact. Then, that point of contact has to push other molecules until all the energy is sufficiently translated into motion of the ball as a whole. So, during that brief moment when the energy is transferring as a wave through the ball itself, I believe the temperature of the ball must rise slightly. If the ball was hit with the same force but prevented from translating that force into motion, the energy wave would reverberate within the ball until it found somewhere else to go. If it had nowhere else to go, what would it become except friction/heat within the ball itself, provided the ball did not express the energy through shape-changes?
 

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