Firedragon52 Posted September 30, 2004 Posted September 30, 2004 I'm not sure whether this question belongs on a Chemistry forum or a Biology forum. But, their are a lot of molecules that make up our body. Loads of H-O and C-H bonds. There is a certain amount of enthalphy (ΔH°*) for destroying (endothermic) and creating (exothermic) bonds. So what is my queston? This is impossible, of course, but say for some reason, every bond in the human body split into atoms, or radicals, or to but it another way, what if a bunch of atoms spontaneously formed bonds in such a way as to create a perfectly functioning human body: By considering the Bond Energys: http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm What do you think would be the total amount of energy in forming the human body? (It should be the same amount of energy as spliting a human body.) Anyone want to take an educated guess?
Martin Posted September 30, 2004 Posted September 30, 2004 I'm not sure whether this question belongs on a Chemistry forum or a Biology forum. But' date=' their are a lot of molecules that make up our body. Loads of H-O and C-H bonds. There is a certain amount of enthalphy (?H°*) for destroying (endothermic) and creating (exothermic) bonds. So what is my queston? This is impossible, of course, but say for some reason, every bond in the human body split into atoms, or radicals, or to but it another way, what if a bunch of atoms spontaneously formed bonds in such a way as to create a perfectly functioning human body: By considering the Bond Energys: http://www.cem.msu.edu/~reusch/OrgPage/bndenrgy.htm What do you think would be the total amount of energy in forming the human body? (It should be the same amount of energy as spliting a human body.) Anyone want to take an educated guess?[/quote'] Just to clarify the question a bit, suppose that instead of say a 70 kilogram human we were talking about 70 kilograms of water. It sounds like you want the energy to disassociate all the molecular bonds so you have hydrogen ATOMS and oxygen ATOMS. do I understand right what you are asking? usually the heats of formation that you find in a handbook, are they not the heat of forming H2O from molecules H2 and (1/2)O2? So you are talking about the energy to divide a body up into isolated atoms of H, O, C, N, with small amounts of Ca, Fe, K, etc. this would take considerably more energy, I would reckon, than dividing the same body up allowing some of the atoms to remain in molecules--- H2, O2, C (graphite), N2,....etc
pulkit Posted September 30, 2004 Posted September 30, 2004 Given that you want to talk in KCal/mol, it would be just a weighted average of all the molecules present in the body. Since many of these are really big, the number I would imagine might be close to 10,000 or 100,000 KCal/mol
Martin Posted September 30, 2004 Posted September 30, 2004 Given that you want to talk in KCal/mol, it would be just a weighted average of all the molecules present in the body. Since many of these are really big, the number I would imagine might be close to 10,000 or 100,000 KCal/mol a strange thought! but I think the average molecular weight of a molecule in the body is quite low, because so much is water I remember seeing in a handbook that the molecular weight of air is about 29. that is the average of around 4/5 being nitrogen and weighing 28 and around 1/5 of the molecules being oxygen and 32. so if you pick a molecule in the air at random, and weigh it, the expected (or average) value is around 29 I think it you pick a random molecule in someone, the expected value of the molecular weight is most likely fairly low----water is 18 and I would not imagine it is all that much larger than 18---not an order of magnitude larger anyway. I see that you disagree Pulkit, well each to his own opinion. ------------------------------ but that leads to all sorts of complications. Pulkit, do you not imagine that the person who asked the question would be satisfied with "kilocalories per gram" or with "kilocalories per kilogram" that would be much easier to estimate one would not have to answer the question of how many grams is a "mole" of human flesh and bone.
Firedragon52 Posted October 1, 2004 Author Posted October 1, 2004 Thanks for the response. I was considering the fact that the human body is made of at least 70% water. But, I figured the rest of the mass would be made up of many organic compounds like lipids, nucleotides, amino acids and carbohydrates.
Martin Posted October 1, 2004 Posted October 1, 2004 Thanks for the response. I was considering the fact that the human body is made of at least 70% water. But, I figured the rest of the mass would be made up of many organic compounds like lipids, nucleotides, amino acids and carbohydrates. you havent gotten even a preliminary answer yet, I am still wondering if you want the answer in terms of "energy per mole" or "energy per kilogram" I can figure out roughly what a kilogram of human consists of, on average. I have no idea what a "mole of human" is, because "human" is not a well-defined molecular species with a known molecular weight. it might be a fun question to consider how much energy is involved in the formation of a person or would be involved in the dissociation of the body into elemental constituents---but we ran into a roadblock with the "per mole" idea. help us get the problem defined and you might get at least the beginnings of an answer
pulkit Posted October 1, 2004 Posted October 1, 2004 Even if the body is 70% water (I did have that in mind while making my earlier posts), the extremely large molecules found else where might still compensate for that. As far as my interpretation of a mole is concerned, you can express as avg molecular weight of a human the total mass / total moles of molecules constituting. Its a defination thats worthless as far as reaching any numerical value is concerned, but seems to be conceptually alright to me. The actual energy of formation of most molecules in the body may not be known, they are generally too complex and occur in too little an amount for any such analysis. It does seem to be an easy problem with any obvious intuitive solutions.
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