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Temperature of something is a measure of the average kinetic energy

  1. May 16, 2005 #1
    If the temperature of something is a measure of the average kinetic energy of its particles then why aren't white water rapids hot?
     
  2. jcsd
  3. May 16, 2005 #2
    I don't think the increase in kinetic energy of a single molecule from being in rapids will be that much compared to the same molecule in, say, a glass. Anyone know the average KE of a water molecule at room temperature/atmos. pressure, etc?
     
  4. May 16, 2005 #3

    FredGarvin

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    If I understand your question, it is due to the rapid mixing of air into the water that causes the white appearance.
     
  5. May 16, 2005 #4

    Astronuc

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    Just because something 'appears' white, that does not mean that something is 'white' hot. White simply means that a sufficient variety of photons are incident upon the eyes so that the color is not red, orange, yellow, green, blue or violet, i.e. the full visible spectrum is represented.

    Ti dioxide is very white - even a room temperature. It simply reflects much of the incident light.
     
  6. May 16, 2005 #5

    russ_watters

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    I think you guys may have misinterpreted: I don't think madness is asking about the white color, but rather the energy of the rapids.

    madness - the energy contained in moving water is insignificant compared to the molecular energy of its temperature. It would take a massive waterfall to measure a difference in temperature from the top to the bottom. You can calculate how much if you like: the energy required to raise a gram of water by 1 degree C is 1 joule. Potential energy is mass times height: 1g of water raised 1000m is 1 joule. So a 1000m waterfall will have a temperature in the basin below 1 degree C higher than in the river above.
     
  7. May 16, 2005 #6
    i am also wqrking mordern techniques on temrature measurements can u help
     
  8. May 16, 2005 #7

    Integral

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    I am not entirely satisfied with this conclusion. I think it over simplifies to much. It totally neglects the fact that there will be energy exchange between water and air. It totally neglects that the velocity of the falling droplet must be transfered to molecular motion, it is not clear to me that this happens with 100% efficiency.

    To understand this we need to understand the difference between the velocity of the center of mass of a droplet and the average molecular velocity of the molecules in the droplet. It is not clear to me that the two are as closely related as your argument requires.

    What is the frame of reference for the average velocity of a molecule? The only answer is that it must be measured with respect to the Center of mass of the body (or droplet in this case). If this is the case then clearly any motion of the C.M. has NO effect on the temperature of the body.


    Consider for a moment the contents of the space shuttle as it is launched into space, does not every atom on board see a large change in velocity? Does the temperature change due to this change in velocity? Sure the skin temp increases due to air resistance but that is an entirely different issue.

    Now, if Russ's argument holds, and 1 deg K is equivalent to a velocity change corresponding to a fall of 1000m, then what would the velocity of a molecule at 273K be? Seems like that is pretty fast. Once again I have reservations about the validity of this argument.
     
    Last edited: May 16, 2005
  9. May 16, 2005 #8

    russ_watters

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    Well, it certainly neglects air resistance, but regarding conversion of the falling droplet kinetic energy to heat, there is nowhere else that it can go. Besides - air resistance is all converted to heat as well.

    I think you may have misinterpreted my point, in any case - I'm not talking about measuring the falling dropplet's temperature (I don't know how or if that would work - I suspect it wouldn't), I'm talking only about the temperature of the basin underneath. When water hits the basin, a lot of things happen (sound, splashes, churning with viscous friction, etc.), but one way or another, all of that kinetic energy becomes heat. It has to, otherwise it would violate the 1st law of thermo.

    The fact that impact energy becomes heat can be seen in such devices as a needle gun or a pneumatic drill. Due to the impacts alone, they get very hot.
     
    Last edited: May 16, 2005
  10. May 16, 2005 #9

    Integral

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    Russ,
    This question is about the temperature of the WATER DROPLET not the rock it hits. Perhaps the impact may increase the temperature of the water as well as that of the rock, that would provide the energy transfer mechanism necessary to increase the velocity of a water molecule. It does not imply that the temperature of the water has increased simply because the C.M. droplet has a velocity.

    Considering the rock, I would guess that most of the energy of the fall can be accounted for in the velocity of the rebounding droplets created by the splash, that energy will then be lost to air resistance.

    I would expect the rock to be in a thermal equilibrium with the water so would expect very little if any transfer of thermal energy between rock and water.

    But, again, the effects of the collision between the rock and the water is not the topic of this thread.
     
  11. May 16, 2005 #10

    russ_watters

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    I didn't mention any rocks, Integral. I'll await clarification from the original poster before going further with this.
     
  12. May 16, 2005 #11

    Gokul43201

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    Temperature is the mean kinetic energy in the rest frame of the body involved. It is defined to be independent of the motion of the body. The energy from the motion of the body is accounted for in its kinetic energy. Saying that its temperature is higher from the motion means you are increasing the thermal energy (internal energy) as well. That is double-counting - you are accounting for the same quantity twice, and that is just bad math.
     
  13. May 16, 2005 #12
    russ_waters,

    That would be quite a coincidence! You meant 1 calorie to raise 1 gm 1 degree.
     
  14. May 17, 2005 #13

    russ_watters

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    Oops - I realize that was a coincidence - I thought the SI system was designed that way. In fact, a calorie is 4.2 joule. Thanks for the correction.
     
  15. May 17, 2005 #14

    dextercioby

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    Russ,that's the definition of calorie.The heat needed to increase the temperature of 1 gr of water from 14.5°C to 15.5°C at normal atmosferical pressure.

    Daniel.
     
  16. May 1, 2009 #15
    Re: temperature

    Am I therefore right in saying that a person sitting astride a molecule of an ideal monoatomic gas would feel the molecule go cold when it enters a Bernoulli type restriction in a tube?
    I say this because the molecule would have lost part of its random kinetic energy, converted into increased velocity of the body along the tube.
     
  17. May 1, 2009 #16

    russ_watters

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    Re: temperature

    No, air does not lose/convert thermal energy when going through a venturi tube, according to Bernoulli. Bernoulli's principle/equation is actually a conservation of energy statement: the total energy of the flow remains constant because while the velocity increases, the pressure decreases.

    So Bernoulli's equation/principle assumes no energy loss in the flow due to friction, viscocity, etc. The OP (and I with my waterfall) was describing a scenario where flow energy is lost. If you wanted to, you could just add an "E" to one side of your conservation of energy statement, but it is extremely difficult to actually calculate the energy - if you knew (ie, if you measured) the energy, you could go back and figure out how it affected the flow.

    This is similar to what an engineer does when they design a pipe, though the tables they use to select the pipe already have that calculated, providing a pressure drop over a certain distance of flow at a certain velocity for a certain pipe. The total pressure loss is calculated and the pump is then sized/selected to provide the necessary flow energy.
     
    Last edited: May 1, 2009
  18. May 1, 2009 #17

    mgb_phys

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    Re: temperature

    Joule supposedly spent his honeymoon in Switzerland trying to measure the temperature difference at the top and bottom of a waterfall.
    In theory the water at the bottom should be hotter because of the loss of PE.
     
  19. May 1, 2009 #18
    Re: temperature

    The molecules lose energy in the radial direction, which explains why the pressure against the walls of the tube decreases. This energy goes into increasning the velocity along the tube.
    Like the man in the spacefraft mentioned above, sitting astride a molecule he would not be aware of this velocity increase. But he would feel the energy drop in the random radial motion. Wouldn't he?
     
  20. May 1, 2009 #19

    russ_watters

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    Re: temperature

    If the spacecraft were equipped with inertial navigation (all are, afaik), he'd be able to measure his change in velocity.
     
  21. May 1, 2009 #20

    sylas

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    Re: temperature

    The mean kinetic energy of molecules in a substance at temperature T in Kelvin is 1.5 kT, where k = 1.38e-23 is Boltzmann's constant.

    Water molecules have an atomic weight of about 18; and so 1 molecule of water weighs 18 / A, where A = 6.02e26; the number of atomic weight units in one kilogram. (Avogadro's number, up by 1000 so I can work in SI units of kilogram rather than gram)

    Suppose you have cold water in the rapids, about 280K. (That's 7 degrees C).

    The mean square of velocity is given by 0.5 mv^2 = 1.5 kT, where m = 18/A.

    Hence v^2 = 3kTA/18 = kTA/6 ~ 400000.

    That is, the molecules in cold water have velocities by virtue of temperature of about 630 m/s. The random bulk motions of turbulent water in the rapids is much much smaller than this. Adding energy to water by agitating it vigorously will increase the temperature, but measuring that is hard.

    Put another way. At any given moment, a large proportion of the water molecules in the Niagara falls are moving upwards.

    Conclusion. There must be something better to do on your honeymoon. Let me think...

    Cheers -- sylas
     
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