Temperature of something is a measure of the average kinetic energy

[madness]

If the temperature of something is a measure of the average kinetic energy of its particles then why aren't white water rapids hot?

 

[El Hombre Invisible]

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?

 

[FredGarvin]

If I understand your question, it is due to the rapid mixing of air into the water that causes the white appearance.

 

[Astronuc]

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.

 

[russ_watters]

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.

 

[efebest]

i am also wqrking mordern techniques on temrature measurements can u help

 

[Integral]

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.

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 transferred 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.

 

[russ_watters]

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 transferred to molecular motion, it is not clear to me that this happens with 100% efficiency.

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.

 

[Integral]

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.

 

[russ_watters]

I didn't mention any rocks, Integral. I'll await clarification from the original poster before going further with this.

 

[Gokul43201]

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.

 

[jdavel]

...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...

russ_waters,

That would be quite a coincidence! You meant 1 calorie to raise 1 gm 1 degree.

 

[russ_watters]

russ_waters,

That would be quite a coincidence! You meant 1 calorie to raise 1 gm 1 degree.

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.

 

[dextercioby]

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.

 

[Roger44]

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. ...

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.

 

[russ_watters]

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.

 

[mgb_phys]

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.

 

[Roger44]

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?

 

[russ_watters]

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?

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

 

[sylas]

If the temperature of something is a measure of the average kinetic energy of its particles then why aren't white water rapids hot?

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

 

[Roger44]

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

Russ, that's not a serious reply. Of course he would also see the walls of the tube flying past him faster than before, but that's not the question. The question is "would the average molecule he's sitting astride feel colder in the Bernoulli restriction?"

 

[Count Iblis]

If you fill a thermos bottle with some cold water (say half full), close it and then shake it vigorously, you could raise the temperature of the water to 100 °C within a few hours. E.g., ignoring heat losses, it would take two hours for half a litre to heat up to 100 °C, if you are able to deliver 30 watts by shaking. Boiling water left in a thermos bottle will typically cool down over a period of 8 hours, so heat loss is not going to prevent one from reaching 100 °C if your arm muscles are not too weak.


The water won't boil because the pressure will always be higher than the vapor presssure. In fact, you could raise the temperature by shaking alone to the critical teperature of water if the thermos bottle could withstand the critical pressure of 220 bar.

As you approach the critical temperature the difference between the water and the water vapor becomes less and less, so heating by shaking becomes less effective.

 

[russ_watters]

Russ, that's not a serious reply.

I'm not sure why you think that - it was meant to be and it responded to the new misconception in the post. Maybe I should have cut the part in the quotes, but...

Of course he would also see the walls of the tube flying past him faster than before...

Ok... but you did state just the opposite in the previous post: you stated that he would not be aware of the velocity increase. I was correcting that misconception.

...but that's not the question. The question is "would the average molecule he's sitting astride feel colder in the Bernoulli restriction?"

...I didn't answer that question in that most recent post of mine because I had already answered it in quite a bit of detail in my previous post. The answer is no.

If there is something in particular about my reply in post 16 that you don't understand, please let me know.

 

[Roger44]

Russ. First off thanks for your patience, and to everybody, sorry to have introduced a second topic which is messing up the waterfall topic :

Russ, you said : "the total energy of the flow remains constant" Yes
"the velocity (along the tube) increases, the pressure (against the walls of the tube)decreases. Yes
"Bernoulli's equation/principle assumes no energy loss in the flow due to friction, viscocity" yes
"The OP (and I with my waterfall) was describing a scenario where flow energy is lost." NO

This is not the case with my man sitting on a molecule. Please consider it to be an idealised Bernoulli case, re-examine the following affirmations, and tell me where my reasoning is faulty :

1. For an ideal monoatomic gas only translational kinetic energy is involved with "heat exchange"
2. The axial velocity along the tube increses, the radial random velocity of molecules must therefore decrease.
3. A man astride a molecule would not be aware of any kinetic energy change thru' increased linear velocity of molecule.
4. He would be aware of lower kinetic energy of molecule thru' reduced random radial motion.
5. He would feel the molecule go cold.

 

[Count Iblis]

In case of an ideal gas you have for adiabatic flow that along flowlines:

1/2 v^2 + h = constant

where h is the enthalpy per unit mass.

h is proportional to temperature, so the temperature goes down as v increases.

 

[russ_watters]

I've written and deleted and rewritten several long responses here and now I'm getting frustrated - maybe I'll come back and finish going over your questions later, but for right now...

I think the issue may be one of dealing with differing principles/versions of Bernoulli's equation. There are more advanced versions of Bernoulli's equation/principle (not developed by Bernoulli) that deal in compressible flow and that may be what you are describing. However, there are a couple of problems here:

1. A waterfall (or white water rapids) has no continuity - no streamlines. It can't, under any circumstances, be considered an application of Bernoulli's principle. Friction, viscocity, and potential energy dominate, and these are not adequatly covered by Bernoulli's principle in any of its forms. In particular, since it is open to the air and static pressure never changes, there can be no compressibility effects (even assuming water was more than a little compressible!).
2. In a venturi tube, air does experience compressibility effects that become relevant above about 220 mph. These compressibility effects result in ideal gas law implications for the flow: change in temperature with pressure/density changes. But these effects are explicitly excluded in the standard form of Bernoulli's equation: the flow is assumed to be incompressible. So when you say:

Please consider it to be an idealised Bernoulli case...

To me, "an idealized Bernoulli case" is the one Bernoulli derived and it explicitly discards the effects you are describing. Some of what you are describing is valid for more advanced versions, though. Read the wiki on Bernoulli's principle, paying specific attention to the descriptions of the two forms they describe: the compressible and incompressible flow forms.
http://en.wikipedia.org/wiki/Bernoulli's_principle

 

[Dadface]

There is an analogy to be drawn between the situation described and the simple free electron theory of electrical conduction.When a current flows there is a low drift velocity which is analogous to the bulk velocity of the falling water and this drift velocity is superimposed on the high thermal velocity of the charge carriers this being analogous to the high thermal velocity of the water molecules.With the waterfall there is a conversion of bulk kinetic energy to thermal energy.Because of the high specific heat capacity of water the temperature rise is not very high and this was first shown experimentally by James Joule.He carried out numerous experiments including measuring the temperature at the top and bottom of a waterfall as mentioned by mgb phys.James took a thermometer on his honeymoon but don't we all?

 

[Roger44]

I think the 5 lines I listed above is a mechanism for cooling in an idealised Bernoulli restriction, true for ANY fluid, compressible or incompressible, which gives and receives heat energy by changes in translational motion.

Thereafter, additional complications arise from compressibility which Russ says produces cooling, and for real gases the Joule-Thomson effect which is cooling for most gases, plus turbulence and friction etc etc

Count Iblis, you say temperature goes down, I suppose you mean the temperature mesured in the frame of the motion, ie the man astride the molecule is holding a thermometer.
But what would be the temperature on the walls of the tube? Can we start by assuming billiard ball energy echanges between wall molecules and fluid molecules.