Regrowing body parts: What are the costs?

Question

Everybody has seen superpowers like superstrength, flying, and regeneration. While flying and superstrength are quite straightforward regarding the energy cost, regeneration is not.

In this question, I want to find the energy cost of this power. Let's assume an arm needs to be regrown. An average human arm weighs 4-6kg. (5kg for the purpose of this question.) Assuming all necessary atoms can be brought to the place they are needed to form an arm one way or the other there is still the energy cost of biochemically forming all the molecules. That is what I want to focus on.

How much energy has to be expended to form an arm from just the materials in their simplest or most common forms? (atoms, very basic gas molecules, etc.)
It can be assumed that H₂O is already available as it is rather common on earth and in a human body. (And certainly more common than H₂ for example)

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Answers should account for as many different costs that occur and thus give an estimation as precise as possible of the total net energy cost of regrowing a limb assuming the required atoms are sufficiently available.

It can be assumed that excess energy of possible exothermic processes can be "gained" and reused for the rest of the energy requirements. (Though I do not think this will be in any case a relevant portion of the net energy cost.)

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This question is a follow-up to my question "Energy cost of creating body parts from nothing but energy". That question showed clearly that the generation of matter by far exceeds the energy cost of biochemically putting a limb together.

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I will post my own calculations as an answer. I am not certain about my math, my data, my conclusions or if I missed a few things, but wanted to include it for other people to improve upon and maybe save answerers some time, research and work.

Answer 1

The data source I used is Composition of the human body.
This clearly shows that proteins, water and lipids make up 97% of the mass of a human body and I will assume that we can get a rather precise estimate if we just account for those to not have to deal with all the other material within the human body. After all this is about getting a good estimate, not a precise calculation.

Water

As stated before we can assume that H₂O is readily available already so we can say there is no energy required to form this. That already accounts for 65% of the mass. So we have a total energy of $$E_{\text{H}_2\text{O}} = 0 \text{kJ}$$

Proteins

I am having serious trouble finding data on enthalpy of proteins. The only thing I found was regarding enthalpy change in protein folding and binding with a value of $$H_\text{proteins}=4 \frac{\text{kcal}}{\text{mol}} \ \ (16.7\frac{\text{kJ}}{\text{mol}}) $$ It is most certainly not referring to the right thing, but to at least have something to calculate with I will use this value, assuming the real value is of a similar magnitude. I put no certainty in that.

The average molecular length of a protein is 375 amino acids for humans. And an average amino acid weighs 110Da (so $110\frac{\text{g}}{\text{mol}}$) This results in an average molar weight of $$M_\text{protein}=41,250 \frac{\text{g}}{\text{mol}}$$. This assumes that the proteins are equally distributed in the body - which is probably false, but might give us a decent enough estimate.

$$m_{\text{arm}} = 5 \text{kg}$$ $$m_{\text{arm;proteins}} = 0.12 \times m_{\text{arm}} = 0.6 \text{kg}$$ $$n_\text{proteins} = m_{\text{arm}} / M_\text{arm;proteins} = 0.01455 \text{mol}$$ $$E = H_\text{proteins} \times n_\text{proteins} - \sum_i {H_\text{material i} \times n_\text{material i}} \ \ \text{(if we assume no byproducts)}$$ $$E_\text{proteins} = 243 \text{kJ} - \sum_i {H_\text{material i} \times n_\text{material i}}$$

I am uncertain what states of the starting materials to assume. Water ($−285.83 \frac{\text{kJ}}{\text{mol}}$), graphite ($0 \frac{\text{kJ}}{\text{mol}}$), nitrogen gas ($0 \frac{\text{kJ}}{\text{mol}}$) might work with possibly negligible additions or taking into account byproducts.

If we were to assume water as a base material we can see the energy needed would increase as the enthalpy of O₂ and H₂ are $0 \frac{\text{kJ}}{\text{mol}}$ of Oxygen would likely be a byproduct in formation of amino acids given this scenario and the enthalpy of formation of water is negative.

Since amino acid structures have about 2-4 Oxygen atoms it is fair to assume that you would need on average 3 H₂O for a singular amino acid. With an average of 375 amino acids per protein $$n_{\text{H}_2\text{O}} = 375 \times 3 \times n_\text{proteins} = 16.369 \text{mol}$$. This gives us with the equation above $$E = H_\text{proteins} \times n_\text{proteins} - {H_{\text{H}_2\text{O}} \times n_{\text{H}_2\text{O}}} \ \ \text{(if we assume no byproducts with H $\neq 0$)} = 243 \text{kJ} - (-285.83 \frac{\text{kJ}}{\text{mol})} \times 16.369 \text{mol} = 4,922 \text{kJ}$$

If we take calorimetric values instead of all I tried to come up with reasonable data, we have 4 kcal per gram of proteins. This results in $$E_\text{proteins;calorimetric} = m_\text{arm;proteins} \times 4 \frac{\text{kcal}}{\text{g}} = 600g \times 4 \frac{\text{kcal}}{\text{g}} = 2400 \text{kcal} = 10,041kJ$$ So about twice as much as I had with the other data.

Lipids

With lipids I am also having troubles finding good data, so I decided to use the calorimetric data which appearantly is between $38,702 \frac{\text{kJ}}{\text{kg}}$ and $39,748 \frac{\text{kJ}}{\text{kg}}$.
I will use $39,748 \frac{\text{kJ}}{\text{kg}}$ as an estimate of the energy used to build lipids, which is probably not correct entirely but might not be off by that much.

$$m_{\text{arm;lipids}} = 0.2 \times m_{\text{arm}} = 1 \text{kg}$$

$$E_\text{lipids} = m_{\text{arm;lipids}} \times 39,748 \frac{\text{kJ}}{\text{kg}} = 39,748 \text{kJ}$$

Preliminary energy sum

Putting all the results together we reach an estiamated total of $$E_\text{total} = \sum_i{E_i} = 44.670 MJ$$ or if we take the calorimetric value for the proteins $$E_\text{total;cal.proteins} = 49.789 MJ$$

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A different result based on this answer by @Freedomjail

This may be a very basic answer but I found this link with some calorific values :

Bone and meat calorific values

This is an approximation but for this

mixture of meat and bone (the composition percentage is not specified) they measured a gross calorific value of 19.69 MJ.kg-1.

This leads to approximatively 112 MJ for your 5.7 kg arm.

From what I recall, the calorific value is the energy gained by burning the material which is pretty similar to separate the molecules into single atoms or at least into some very smaller molecules. If so, this is the opposite of the energy needed to creates molecules from single atoms.

Now I was talking about a 5kg arm so taking that into account that would make 98MJ.

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So we got estimates ranging from 44MJ to 98MJ.
According to a calorie calculator my recommended daily intake is 8140kJ (8.14MJ). This means the energy required would be 5.4-12 days worth of calorie intake.

Answer 2

I'll put my sources at the bottom of the piece:

That 5kg of arm is going to need 3250g of Oxygen, 900g of Carbon, 500g of Hydrogen, 150g of Nitrogen, and the rest is fiddly non-gases that'll need to be sourced elsewhere

We can atmospherically source most of that, but we'd need to strip about 7500 m³ of dry air to get enough hydrogen so if we assume 1% water vapour, about average at sea-level, then we can get away with taking some or all of the gases from only about 35m³ or 35,000 litres of air to get all the Carbon we need, we can get the Oxygen etc.. from that volume with relative ease.

We'll need to hit the ground, plants, and animals around us up for another 200g of solid elements, mostly Calcium and Phosphorus, that could be pretty messy but there you go.

Based on pigs, because they have about the same body chemistry and funnily enough no-one has this figure for people, we normally convert about 35% of what we eat into tissue, the end caloric value of a 5kg arm is about 23,500 food calories so we'd need 67,000 calories to support it's production. I make that 280MJ but that does assume a hugely wasteful natural metabolism, if we use more efficient direct conversion then your math is about right and it's only about 50MJ, a lot but not appallingly so.

Alternatively if there is an arm composed of everything we need to regrow it, already packaged up as proteins etc... lying on the ground, or otherwise detached but not yet at any great distance from the hero. Like in the bad guy's unsuspecting hand, procurement and processing gets a whole lot cheaper.

Composition of the body

Atmospheric composition

Atmospheric density

Food conversion rates