Asklemmy

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/

Speaking only in terms of energy:

Molecules attract (adhesive and cohesive) causing capillary rise. Actually pottential energy due to that force is reduced converted to kinetic energy of motion which then gets converted into the gravitational potential energy mgh.

Now capillary rise won't happen endlessly. It stops at a certain point where it cannot pull more water.

You could evaporate this water so that more water would flow up(also works in trees, but their mechanism of pulling water is more sphisticated). But now, on evaporating, you are applying more energy to it, which molecules held by adhesive and cohesive forces are puller apart, making it gain more potential energy to pull more water from bottom.

The reason why it rises is to minimise the potential energy and it does not increase energy

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

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.

∆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.

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.

I don't remember all of the details, but I thought it was essentially the water's surface tension that foots the energy bill when climbing a paper towel or a capillary in a tree.

The surface of fluids like water are unhappy. Molecules on the surface would much rather be deep in the fluid because on the surface they have "dangling" Van der Waals & polar bonds to one side. You can calculate the potential energy of the surface due to all of those dangling weak bonds, & that's the energy that is used to climb a capillary (the energy isn't free).

I could be misremembering though, I admit. School was many years 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

I've always wondered this too.