The presence of iron micro spheres in the dust collected from 911 collapses is considered the “smoking gun” for controlled demolition. While this site is not a de-bunking site, it is impossible to avoid many of the arguments put forth by CD proponents. Here is the flaw with iron micro spheres.
It is true that round spheres of iron can only be produced in an environment of liquidification, which requires heat. The leap is then made that steel must have melted in order to create the micro spheres. Melted steel means temperatures not attainable in the fires of 911, which can only mean thermate, the incendiary proponents claim was used to cut apart the structural members of the buildings.
The flaw is assuming steel must have melted. No where do the cited scientists consider that rust reverts to iron at it’s boiling point of 300c (572f), well within the temperatures achieved in all 3 buildings on 911.
Structural remains show considerable rust in all 3 buildings, and heavy rust in some areas of the Twin Towers. Failure to properly consider all sources for the micro spheres casts a dim light on the research. The “smoking gun” just became a water pistol. No grand jury will buy it.
The source link is no longer valid, but here is what I copied and saved of the article:
300 c is boiling point for rust.
Celsius to Fahrenheit: (deg Celsius x 9/5) + 32 = (300 x 9/5) + 32 = 572 deg Fahrenheit
Water freezes spontaneously below 0oC, and ice spontaneously melts above 0o C. …and oxygen forms rust, but rust does not spontaneously change back to iron. … the temperature at which the process is reversible is the normal boiling point.
Walter S. Hamilton, Ph.D
Entropy and The Second Law of Thermodynamics
For a chemist, one of the main reasons for studying thermodynamics is to be able to predict whether or not a reaction will occur when reactants are added together under a given set of conditions (temperature, pressure and concentration). A reaction that does occur under the given set of conditions is called a spontaneous reaction. If a reaction does not occur under specified conditions, it is said to be nonspontaneous. The following few examples are spontaneous chemical and physical processes that can be observes in everyday life:
Water runs downhill, but never up, spontaneously.
A lump of sugar spontaneously dissolves in a cup of coffee, but dissolved sugar does not spontaneously reappear in its original form.
Water freezes spontaneously below 0oC, and ice spontaneously melts above 0oC.
Heat flows from a hotter object to a cooler one, but the reverse never happens spontaneously.
Iron exposed to water and oxygen forms rust, but rust does not spontaneously change back to iron.
These examples show that processes that occur spontaneously in one direction cannot, under the same conditions, also take place spontaneously in the opposite direction.
From a study of these examples and many, many others we conclude that being exothermic favors a reaction or process being spontaneous, but does not guarantee it. In other words, we cannot decide whether or not a chemical reaction will occur spontaneously solely on the basis of energy changes in the system.
In order to predict the spontaneity of a process, we need to know two things about the system. One is the change in enthalpy, the other is entropy (S), which is a measure of the randomness or disorder of a system. The more disordered a a system, the larger its entropy. Like enthalpy and internal energy, entropy is a state function. The change in entropy of a system, ΔS = Sfinal – Sinitial, depends only on the initial and final states of the system and not on the particular pathway by which the system changes. A positive ΔS indicates an increase in randomness or disorder. The entropy change of a system is defined in terms of the heat transferred to the system during a reversible path from initial state to final state, regardless of how the process actually takes place. We refer to this heat transferred as qrev where the subscript ‘rev’ means that the path between the states is reversible. If a process occurs at constant temperature, the entropy change is defined as qrev divided by the absolute temperature:
ΔS = qrev / T (Constant T)
An example of a reversible process that occurs at a constant temperature is a phase change at the temperature at which the two phases are in equilibrium, such as the boiling of water at 100oC.
Calculate the entropy change when 1 mol of water is converted to 1 mol of steam at 1 atm pressure.
The amount of heat transferred to the system during this process is the heat of vaporization, ΔHvap, and the temperature at which the process is reversible is the normal boiling point. For water ΔHvap = +40.67 kJ/mol and Tb = 100oC = 373.2 K.
ΔHvap (1 mol)(+40.67 kJ/mol)(1000 J/1 kJ) ΔSvap = ----- = ----------------------------------- = 109.0 J/K Tb (373.2 K)
We are now in a position to discuss why certain processes are spontaneous. The connection between entropy and spontaneity is expressed by the second law of thermodynamics: The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Since the universe is made up of the system and the surroundings, the entropy change in the universe for any process is the sum of the entropy change in the system and the entropy change in the surroundings.
For a spontaneous process: ΔSuniv = ΔSsys + ΔSsurr > 0
For a equilibrium process: ΔSuniv = ΔSsys + ΔSsurr = 0
Entropy Changes in the System
To calculate ΔSuniv we need to know both ΔSsys and ΔSsurr. First lets focus on ΔSsys. The standard entropy values of a large number of compounds have been measured in J / K · mol. To calculate ΔSorxn (which is ΔSsys), we look up their values in a table and use the equation:
ΔSorxn = n So(products) – m So(reactants)
This equation is similar to the one we have used to determine the enthalpy of a reaction. Now we will use the second law to decide if the reaction for the synthesis of ammonia is spontaneous at 25oC.
N2(g) + 3 H2(g) ==> 2 NH3(g) ΔHo = -92.6 kJ
use the equation above to get ΔSorxn or ΔSsys
ΔSorxn = [2 So(NH3)] – [So(N2) + 3 So(H2)]
= (2 mol)(193 J/K · mol) – [(1 mol)(192 J/K · mol) + (3 mol)(131J/K · mol)]
= 199 J/K
We can calculate ΔSsurr from the following relationship.
-ΔHsys -(-92.6 x 1000)J ΔSsurr = -------- = ---------------- = 311 J/K T 298.2 K