Info

Radiation Efficiency

Fig. 5.4. Dependency of hEx on radiation efficiency coefficient, t]r, for two different values of K(K\ , K2).

"information work". It is easy to see that K ! 0 then HRr)i ! 0, and the curve HEx has only the "classic" branch. In this case the system works only like a classic thermodynamic machine. Conversely, if K n then (hR)i ! 1, the curve hex has only the "non-classic" information branch. The system works purely as an information machine producing an infinite amount of information. These cases are marginal; substantively the active surface is a composition of these two ideal machines and it produces both mechanical (or chemical) work and information.

It is interesting that on the left-hand branch the exergy efficiency increases with a decrease in the radiation efficiency, i.e. with a decrease in energy absorbed by the active surface. At its limit, where the absorbed energy tends to zero, the exergy efficiency tends to a non-zero value, which is equal to K. Moreover, on the left-hand branch the exergy is always more than the radiation balance. If we keep in mind one of the possible interpretations of exergy as the maximal useful work which can be performed by the system, and consider the radiation balance as a portion of external energy consumed by the system, then, formally, we have a situation when the low energy conservation is not fulfilled. At first sight, this is a paradoxical result, but it is an artefact of such a representation of exergy and our assumption about the independency of Hr and K. Indeed, if we write the expression for partial exergy as Ex,- = E°ut ln(Eout/Ein) + (Ein — Efut) then it is just as easy to see Ex,-! 0 for Eout ! Efut; therefore, the variables hR and K are not really independent, but the closer hR is to zero, the more they are dependent. In other words, the model is not completely adequate in the vicinity of hR = 0. Nevertheless, the model is working even on the left-hand side. This is explained by the fact that the information can be transferred even where there is a very weak signal. The point is that the information measures are functions of frequencies, but the frequency pi = N/N may tend to a finite value even if N (and N,, of course) tends to zero. This, by the way, explains the fact that even a weak signal can transfer a relatively large quantity of information.

Conversely, as we can see in Fig. 5.4 on the right-hand branch, when hR ! 1, the value of HEx does not depend on K (two branches, corresponding to different values of K, come together when hR ! 1). This is natural since the main performance of the classic machine is not to produce information but only to do mechanical or chemical work.

5.7. How to calculate the exergy of living organic matter?

If we compare the living and dead organic matter then we see that their chemical compositions almost do not differ from one another; moreover, their thermodynamic characteristics such as enthalpies are identical. The same statement is true for the biomass of all living organisms (see Chapter 2). However, if the detrital organic matter is a mixture of carbohydrates, fats and proteins then the living organic matter is an ordered structure created from these materials in accordance with a plan given by genetic information. Naturally, both the biomass of living organisms and the mass of detritus could be measured in the same units, since 1 g of biomass and 1 g of detritus have the same chemical composition, but their roles in the ecosystem differ significantly from each other.

Therefore a question arises: what is the difference between the same quantities of detritus and living matter? We postulate that they differ in the exergy contents.

The problem is how to calculate the exergy contained in 1 g of living organic matter? We assume that the process of creation of living matter can be represented as the chemical formation of organic molecules from non-organic substances of the environment and simultaneously the ordering of these molecules in such structures as organisms in accordance with their genomes. As a result the specific exergy of living matter (per 1 g) will be ex = exchem + exbio' = T0(infchem + infbid) (7.1)

where exchem is the specific exergy obtained by living matter due to chemical processes, and exbl°l is the specific exergy as a result of ordering, i.e. the construction of an organism in accordance with genetic plan. In other words, we have to find the correct composition of the enzymes determining the biochemical processes in the organisms. Since exergy has an information interpretation, ex can be represented as a sum of two information quantities multiplied by the temperature of the environment, T0: the first is the content of "chemical" information and the second is the content of genetic information in 1 g of biomass.

Since there is no difference between the chemical compositions of dead and living organic matters, the first item in Eq. (7.1) can be associated with the exergy of detritus. But how to calculate this? At first sight the problem seems very complex, but as Anton Chekhov said: "The gun hanged on the wall in the first act has to fire in the last act". We have "hung our gun" in Section 2.7 of Chapter 2, where we calculated the standard Gibbs' potential for detritus, AG0. Now it is time to "fire it". If keeping in mind that the exergy accumulated by the system is equal to the increase of its Gibbs potential then exchem = A- AG0 = 18.7 kJ/g . (7 . 2)

The appearance of the factor (T0/300) is explained by the fact that AG0 was calculated at the temperature of 300 K. Since the exergy can be represented in the form exchem = 300 AG0 = T0 infchem = T0 62.3 J/Kg . (7 . 3)

the quantity of information contained in 1 g of partially ordered organic matter of detritus and expressed in the units of thermodynamic entropy is equal to infchem = 62 . 3 J/Kg . (7 .4)

It is known that the same information, expressed in bits, infbhtem, will be equal to infbhtem = log2 wd = 1. 44 ln wd where wd is the number of possible chemical states of 1 g of detritus. The problem is how to establish the similarity between infchem and infjbhtem? Here we are compelled to use the thermodynamic model of ideal solution, i.e. to consider detritus as an ideal solution of organic matter in water. It seems fully unrealistic, but keeping in mind that even less realistic models in physics and chemistry nevertheless give good results, we think that this model should be sufficient. Then

¡nfchem = 62.3 J/K g = wd = -r3RMdinf'''em <7 5>

where R = 8 . 4 J/Kmol is the gas constant and Md < 105 is the molecular weight of detritus. Using this relation we can estimate the weight of 1 bit in the units of thermodynamic entropy: 1 bit = R/1 . 44Md < 5 . 8 X 10~5 J/K g.

The genetic information is calculated very easily (see Section 4.9, Chapter 4). Living organisms use 20 different amino acids and each gene determines on average the sequence of about 700 amino acids. Therefore, the information infbi° can be found from the number of permutations among which the characteristic amino acid sequence for the considered organism has been selected (inf^ = /nuc of Section 4.9):

ifbit = 1. 44 ln(20700gi) = (1.44 X 700 ln 20)gi < 3000g bits (7 . 6)

where gi is the number of non-nonsense genes in the genome of the organism of some ith species or some taxon (see Table 5.1). Using the value of 1 bit expressed in entropy units we get infbi°l < 5.8 X 1025 X 3000gi = 0.174gi J/K g. (7.7)

Table 5.1

Numbers of non-repetitive genes in genome (gi) and weighting factors (ft) for living organisms (J0rgensen et al., 1995a,b, 1998, 2000)

Table 5.1

Numbers of non-repetitive genes in genome (gi) and weighting factors (ft) for living organisms (J0rgensen et al., 1995a,b, 1998, 2000)

Organisms

gi

ßi

Detritus

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