I've randomly been thinking about smoothies and internal energy all weekend. If we have an assortment of fruit in the solid phase and then proceed to blend it all so that it ends up being in the liquid phase, am I right in saying that work has been done on the solid fruit such that its internal energy is higher, therefore it remains in the liquid phase at the same temperature as the room (once thermal equilibrium is reached). I'm having a bit of difficulty in understanding how another identical assortment of solid fruit and the smoothie liquid can both be at equilibrium if their kinetic energies are different (liquid has more than a solid right?). I thought that kinetic energy determines temperature??


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Well, is fruit really in a solid phase? Consider that fruit consists of a lot of water; that water is in a liquid phase prior to blending. Saying that fruit is a solid is like saying a water balloon is solid.

When you blend the fruit, all you're really doing is slicing it up. What's left is fragments of fibres, membranes and so on in a liquid-ish suspension because you also freed up water in the process.

Of course the blending itself adds some energy but this is negligible.

To answer your question on a single line: There's no difference in molecular kinetic energies or temperatures because there are no phase transitions on a molecular level.

  • $\begingroup$ Thank you for that. Yes that makes sense to me now. What about a glass of water and water vapour together at thermal equilibrium in a room? The vapour has more kinetic energy right? Yet they are the same temperature. Wouldn't there also be more potential energy in the vapour? So more total energy as a whole and yet it is at the same temperature? $\endgroup$ Feb 8, 2014 at 21:04
  • $\begingroup$ Water vapour as in fog or clouds is just a bunch of microscopic water droplets. In a true gas phase, evaporating requires a certain amount of energy, you could think of it as the gas 'holding' potential energy. $\endgroup$
    – Kvothe
    Feb 8, 2014 at 21:25

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