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this is my first post here. I watched this video https://www.youtube.com/watch?v=h5wQoA15OnQ and at 51 seconds it describes Bernoulli's effect of a drop in pressure resulting from an increased velocity (when the liquid enters the smaller capillary tubes). Then it states that the refrigerant (the liquid) boils because of this drop in pressure which lowered the boiling point AND there is a resulting temperature drop in the liquid.

Ok so when a liquid boils due to a drop in pressure (like in the throttling device of a refrigerator) why is there a loss of heat, specifically how does the liquid lose its own energy upon evaporation? I'm looking for an explanation that is deep and explains the mechanism from a molecular level.

When the liquid evaporates my understanding is that there is a gain in kinetic energy of the molecules that are escaping and because this energy was present to begin with (because no energy was added) there must then be a loss in the kinetic energy of the molecules in the liquid so the temperature OF THE LIQUID drops.

Ok maybe i just answered my own question but does this then mean that a gas of a fluid has a lower heat capacity than its liquid counterpart, which would be why the gas form of the refrigerant can get rid of heat (to the external environment (in this case away from the refrigerators insulated inside)) more efficiently, or maybe this isn't the case but rather the heat is now more concentrated (is mostly in the gaseous portion of the fluid) and so it can then be released more efficiently...

I'm just unclear on exactly how there is heat loss from a liquid boiling due to a drop in pressure.

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  • $\begingroup$ You'll note that the video doesn't actually mention Bernoulli's principle at all (for good reason). In this case the pressure drop is mostly due to energy dissipation (viscous losses in a narrow pipe). Potential energy (pressure) is primarily being lost, the amount converted to kinetic energy is minimal. $\endgroup$ – user3823992 Jun 21 '15 at 3:29
  • $\begingroup$ Is viscous loss in a narrow pipe mutually exclusive from Bernoullis principle (can there be viscous loss in a narrow pipe and not be any effect of Bernoullis principle). I looked up the viscosity of a common refrigerant (HFC-134a) and it is less viscous than water. www2.dupont.com/Refrigerants/en_US/assets/downloads/… $\endgroup$ – Daniel Jun 21 '15 at 22:29
  • $\begingroup$ Bernoulli's equation comes from the conservation of energy. When viscosity is significant energy is not conserved, so the equations become less accurate as viscous effect decrease. In the video, you see that the tubes expand back to their original diameter after the tube. If Bernoulli's equation applied, the flow would return to the original state. $\endgroup$ – user3823992 Jun 22 '15 at 15:46
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Any form of evaporation will cool a liquid, whether it's evaporation due to boiling or evaporation at temperatures below the boiling point.

The temperature of a liquid is related to the average energy of the molecules that make it up. However at any moment in time the energies of the molecules are not all the same - some will have less energy that the average and some will have more energy than the average. The energies of the molecules will follow something like a Boltzmann distribution.

It's always the molecules with the higher energies that boil first, obviously because they are the ones with enough energy to escape from the liquid into the gas. But if we remove the molecules with the highest energies then this must reduce the average energy of the remaining molecules, and therefore the temperature falls.

That's why any form of evaporation cools a liquid - because it preferentially removes the molecules with the highest energies.

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  • $\begingroup$ I understand that but in this case when evaporation occurs in the throttling device the molecules don't just float off but they are still part of the fluid that will be used to absorb heat from the inside of the refrigerator. So this system must consist of both the liquid and gaseous parts of the fluid because the fluid is in a capillary tube and heat is the only thing being exchanged with the environment so how upon evaporation is the fluid in its entirety able to lose energy and what is special about the case of the fluid boiling because of a drop in pressure and not an addition of heat? $\endgroup$ – Daniel Jun 20 '15 at 18:45
  • $\begingroup$ @Daniel, the heat to boil the liquid comes from the liquid itself. That is why the liquid temperature drops. Also, see my original answer on this subject. $\endgroup$ – David White Jul 17 '16 at 23:01
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There is a misnomer in the original question. There is NO loss of heat when the liquid pressure drops and the liquid boils at a lower temperature and pressure. Energy is conserved, so the high pressure, high temperature liquid, boils into a low pressure low temperature mixture such that the heat content of the low pressure liquid plus the heat content of the low pressure vapor equals the original heat content of the high temperature liquid. Whether or not the high temperature liquid will boil is determined by the Antoine equation, which mathematically ties together the vapor pressure vs. temperature of the liquid in question.

Just keep in mind: heat is NOT temperature. Heat is energy, which can exist at ANY temperature above absolute zero.

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