# Tag Info

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The key point I'm getting at is that when the pressurized liquid moves through the throttling valve, the auto-refrigeration effect is really a way of splitting the hot vapor "part" away from the cold liquid "part". I think this is the main misconception you have. Typically when a material boils, the gas that is released is at roughly the same ...

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It depends on how you define your system or your control volume. If only the container is considered then indeed the entropy has decreased due to cooling. On the other hand if you account for the container plus the escaped vapour the entropy has increased, as the randomness of the molecules in the vapour state is larger than compared to in liquid state at ...

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No, in fact you could even view the spontaneous evaporation as being driven by the fact that it increases entropy. Basically what's happening is the liquid particles have random speeds (with distribution characterized by temperature), and they bump into each other. Every once in a while, two particles near the interface will collide in just such a way that ...

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Temperature is a macroscopic concept, so you're bound to run into some problems when you apply it on a molecular scale (what does temperature and equilibrium (or for that matter, friction) even mean on such a small level?). A thermal equilibrium does not mean all the molecules have the same energy. The distribution of their energies looks like this (normal ...

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Put the ice cube atop. Poorer air and water conductive and convective heat transfer kinetics are more than compensated by meltwater and cooled air dribbling down the pot, further cooling it by heat capacity as they warms from 0 C. If the ice cube is touching the bottom, meltwater and cooled air flow away from the pot.

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Open or not open, touching or not touching, ice cube above is better than ice cube below where cooling of the pot is concerned. Your choice is correct but it is more accurate to say that air cooled by the ice cube sinks onto the teapot thereby cooling it (note that this mechanism cannot occur if the cube was under the teapot). This isn't a question about ...

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If the Reynolds number of laminar flow exceeds about 4000, it will go turbulent. Turbulent flow will choke mass transfer, and worse if it a supersonic shock. Rate of heat transfer through source and sink walls is another limit. And if you really juice it (accelerator)...the beta factor in special relativity. That last can be real world - ion engines in ...

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There are basically three limitting factors in the system you set up. Both evaporation as well as condensation require in practice som sort of liquid films, which provide thermal resistance between the walls and the vapour phase and therefore temperature gradients are required to allow for heat transfer. This means that besides geometry and fluid dynamic ...

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First figure out how you should solve it with a psychrometric diagram $(p_w;T)$: Find the wet bulb temperature on the saturation curve; Follow isenthalpic curve until reaching the dry bulb temperature: you have found the mixture point; Then follow the isobaric curve until reaching the saturation curve: you reach the dew point temperature. Now, knowing ...

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