For example in a tree, the water is lifted from the high concentration in the soil to the low concentration higher up in the tree. But at the end of that process the water has been elevated, which should take energy (=mgh), but it seems like it kind of gets lifted for free without spending any energy?

Similarly, dipping a paper towel into a bowl of water, the water “climbs” the towel (by capillary action?) and absorbs upwards, meaning the water was lifted upwards (so gained potential energy) seemingly for free?

  • TheMetaleek@sh.itjust.works
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    7 hours ago

    As others have stated, water in trees gets up thanks to two processes. The first is indeed capillary action. The tubes carrying the water are rather thin, and it clings to the sides of it. But this is a rather small part of the total energy carrying the water. The main mechanism is a negative pressure inside the vascular system of the tree. Basically, tree leaves sweat water all the time (more or less depending on temperature). The water leaving the tree kind of sucks up the water following inside the vessels (this is a simplification to not go into the physics behind). In some larger trees, the negative pressure inside the vascular system can be exceptionally strong, requiring exceptional strength of the tree’s components.

  • Bilbo_Haggins@lemm.ee
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    9 hours ago

    What you are describing is not osmosis, it is capillary action. Capillary action is caused by the forces between the water molecules and the molecules of the tube overcoming the force of gravity. You can read more here: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Liquids/Capillary_Action

    Briefly, the water molecules are attracted to the molecules of the tube by adhesive force. The liquid molecules are also attracted to each other by cohesive force. The interplay of these forces causes capillary action.

    However, it seems that tree sap moves by more than just capillary action. If you scroll down part way in this book they talk about it a bit: https://pressbooks.online.ucf.edu/phy2053bc/chapter/cohesion-and-adhesion-in-liquids-surface-tension-and-capillary-action/

  • petl
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    9 hours ago

    No energy is required for regular osmosis as it is a statistical proces involving the random movement of particles: https://www.sainaptic.com/post/what-are-the-differences-between-diffusion-osmosis-and-active-transport

    Trees have a vascular system for water transport: https://www.earthdate.org/episodes/how-trees-lift-water

    The energy required for a piece of paper to get wet upwards is provided by the reduced surface energy: https://physics.stackexchange.com/questions/2254/when-water-climbs-up-a-piece-of-paper-where-is-the-energy-coming-from

  • bananaslug4@lemmy.blahaj.zone
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    9 hours ago

    ∆G = ∆H - T∆S

    ∆G is the change in Gibb’s free energy. If it is negative for a process, the process will happen spontaneously. If it is 0, the process is at equilibrium. If positive, the process will not occur unless coupled to another process to make total ∆G ≤ 0.

    ∆H is the change in enthalpy, the heat energy of the process. If it is negative, the process releases heat to the environment, getting hotter. If positive, it absorbs heat from the environment, becoming colder. If that feels counterintuitive, remember that you as the observer are also the environment.

    T is temperature in Kelvin.

    ∆S is entropy. Entropy is hard to rigorously define, but loosely it represents a state of disorder. A well mixed solution has high entropy, since the degrees of freedom within the mixture are high. A concentration gradient (high salt on one side of a membrane, lower on the other) has lower entropy because the existence of that gradient restricts those degrees of chemical freedom. A good rule of thumb is that if a barrier is required to maintain a state of things, it is a lower entropy state than what is possible.

    Put that all together, and we can think about the question again. ∆H is close to zero for the process. It will do some slight cooling, but that has more to do with evaporation than anything else. Temperature is unknown but doesn’t affect the sign of ∆G if ∆H is close to zero. That means ∆S is our main driver. In the case of a plant, there is a gradient, with more salt inside the root than outside it. As such, in order to increase entropy and therefore have a negative ∆G, water moves from a low-salt environment to a high-salt environment. This brings water into the root and in doing so creates water pressure that forces the water upward as long as it has a path to do so.

    The logic is similar but simpler for a piece of paper sucking up water, as the gradient is caused by the paper being dry and therefore creating a gradient in the amount of water.

  • wewbull@feddit.uk
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    9 hours ago

    I don’t know, but given that forces involved are on the molecular level, I suspect those are driven by the kinetic energy of particles in a fluid (i.e. heat) and that there’s a very slight cooling going on.

  • dysprosium@lemmy.dbzer0.com
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    9 hours ago

    Not sure but I think because water sticks to surfaces and pulls on other water molecules. I think this is what the capillary effect is based upon. Thus also (partly) how trees get their water upwards, and how sponges absorb water