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Just out of curiosity, I'm trying to get a sense of the order-of-magnitude theoretical thermodynamic cost of food absorbtion. I'm thinking of food absorption as "move nutrients in newly consumed food to their desired location in the body". I am just wondering if this way of estimating it is broadly correct.

  • Assume that a piece of food enters the small intestine from the stomach, and approximate it as a fully mixed ideal gas consisting of all the various nutrients.

  • Assume the nutrients have already been broken down into their constituent parts, and the cost of this is zero.

  • Assume each nutrient has to move to some particular location in the body (possibly spread out over all the cells in the body). Assume this is thermodynamically equivalent to simply separating the nutrients in the food (once they are separated, they can be moved to their target locations without decreasing their entropy).

Using these assumptions, the thermodynamic cost is simply the cost of separating mixed ideal gases of nutrients. Let's assume we just have 3 nutrients and water, in equal proportions (glucose, fatty acids, amino acids, and water, each 25% of the volume), and lets assume 300 Kelvin. Then we get that the energy cost of separating 1 mole of each of these ideal gases is given by $\Delta E=300*N_A*k_B*log(4)\approx 300*6*2J=3600J=3.6kJ$.

(E.g. one mole of glucose weighs 180 grams, which has a nutritional content of about $3000kJ$, so the lower-limit thermodynamic cost of "absorbing" glucose from food under these assumptions is about 0.1% of the energy content of that glucose.)

Is this kind of reasoning basically right, for an order-of-magnitude estimate? Or am I missing something fundamental about digestion and so on?

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  • $\begingroup$ en.wikipedia.org/wiki/Adenosine_triphosphate "The energy used by human cells in an adult requires the hydrolysis of 100 to 150 mol/L of ATP daily, which means a human will typically use their body weight worth of ATP over the course of the day.[26] Each equivalent of ATP is recycled 1000–1500 times during a single day (150 / 0.1 = 1500),[25] at approximately 9×1020 molecules/s.[25]" $\endgroup$
    – The Tiler
    Commented Jun 25, 2023 at 9:18
  • $\begingroup$ @HVAC, this is interesting but could you give a hint as to the connection to my question? $\endgroup$
    – user56834
    Commented Jun 25, 2023 at 14:24
  • $\begingroup$ "Assume the nutrients have already been broken down into their constituent parts, and the cost of this is zero". "Using these assumptions, the thermodynamic cost is simply the cost of separating mixed ideal gases of nutrients" " Using these assumptions, the thermodynamic cost is simply the cost of separating mixed ideal gases of nutrients". sorry, I did some biology and geology at university and I did not understand anything because the hypotheses are contradictory. You have to specify what you are looking for: the energy of transport or separation. $\endgroup$
    – The Tiler
    Commented Jun 30, 2023 at 11:40
  • $\begingroup$ By hypothesis the separation energy is zero :"Assume the nutrients have already been broken down into their constituent parts, and the cost of this is zero". $\endgroup$
    – The Tiler
    Commented Jun 30, 2023 at 11:48
  • $\begingroup$ @HVAC, by separation I mean, un-mix. The nutrients are mixed together in a single fluid (or in my assumptions, ideal gas). Separating them into separate volumes of nutrient has thermodynamic cost. By separating I don't mean breaking nutrient molecules down into constituent parts, e.g. breaking down carbohydrates into glucose, I'm assuming THAT has already been done. $\endgroup$
    – user56834
    Commented Jul 1, 2023 at 5:20

2 Answers 2

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I think it is hard to define the cost of food absorption. Overall, the food absorption results in release of energy, rather than expending it, although on some stages energy should be spent in order to absorb food (and subsequently release energy) - e.g., chewing and swallowing are energetically consuming.

Metabolism is basically decomposition of high energy hydrocarbons (like glucose or fatty acids) into lower products - mostly $CO_2$ and water. In other words, it is a chemical reaction, in which complex molecules are split into elementary one. (Note also that the reaction path is not that of a burning reaction, even if the products are similar - see Can we call rusting of iron a combustion reaction?)

The chemical reaction however happens in all the cells of the organism - i.e., the hydrocarbons have to be distributed to all the cells. Many organism do actively help this process externally - via hunting, cooking, chewing, swallowing - and internally - via contractions of the digestive system and the blood circulation. However, there are also many organisms that rely purely on diffusion and surface tension - notably plants. One could probably argue that these also play the essential role in distributing the nutrients to cells in all organisms.

Thus, if as the OP suggests, the food is already decomposed in its basic elements, then moving it costs next to nothing. A caveat is that we need to have non-equilibrium conditions, which force the movement, such as:

  • excess concentration of nutrients at one site, so that they can subsequently diffuse
  • heat, which may cause body liquids circulating (e.g., plants and cold-blooded animals)
  • gravity (in many animals) or natural air/water flows (e.g., in case of sponges)

Related:
Life and Death, and Energy Conservation
Should fat calories' numerical values be increased to better reflect the new scientific findings?
Does diffusion do work?

Also, Mathematical Biology by Murray, although it does not contain answer to this question, is a good text for anyone interested in physicists view of biological processes.

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  • $\begingroup$ "However, there are also many organisms that rely purely on diffusion and surface tension - notably plants" But then the nutrients still need toend up in the right location within a cell don't they? e.g. connected to a particular receptor type on a cell membrane or stored in some capsule. Isn't there still some thermodynamic cost associated with that, since it decreases entropy relative to the nutrient just floating around fully mixed with other nutrients in the blood? Perhaps this thermodynamic cost is paid in setting up the receptor, and released once the receptor captures the molecule. $\endgroup$
    – user56834
    Commented Jun 30, 2023 at 7:44
  • $\begingroup$ @user56834 One has to define unambiguously the thermodynamic cost that you are talking about. A cell is like a heat engine, which takes energy from one reservoir, does some useful work and rejects the unused energy. What is thermodynamic cost of a heat engine? Of a Carnot engine? There is some intuitive notion behind it, but one cannot answer the question, if precise physical definition is not given. $\endgroup$
    – Roger V.
    Commented Jun 30, 2023 at 7:47
  • $\begingroup$ Aquaporins are responsible for the high water permeability of membranes. en.wikipedia.org/wiki/Aquaporin , it also has pumps that work with the presence of the Na+ ion to bring in the glucose... $\endgroup$
    – The Tiler
    Commented Jul 1, 2023 at 16:26
  • $\begingroup$ The Na+/K+ and Ca 2+ pumps as nutrient separators, so they consume energy. en.wikipedia.org/wiki/Sodium%E2%80%93potassium_pump en.wikipedia.org/wiki/Calcium_ATPase $\endgroup$
    – The Tiler
    Commented Jul 1, 2023 at 20:51
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So not all cellular transport (separation:extracellular to intracellular or vice versa) mechanisms are energy consuming, there are passive mechanisms:

Active transport is the movement of a substance across a membrane against its concentration gradient. This is usually to accumulate high concentrations of molecules that a cell needs, such as glucose or amino acids. If the process uses chemical energy, such as adenosine triphosphate (ATP), it is called primary active transport. Secondary active transport involves the use of an electrochemical gradient, and does not use energy produced in the cell (*)

(*) https://en.wikipedia.org/wiki/Membrane_transport_protein

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