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The principle of heat release regarding phase change latent heat

2016-04-29View Original

Thread Content

H2O(g) + H2O(g) = 2H2O + 0.8eV (1.54μm photon, equivalent to blackbody radiation at 1604°C, in a metastable gaseous state) 2H2O + H2O = 3H2O + 0.6eV (2.1μm photon, equivalent to 1103°C, in a metastable state) 3H2O + H2O = 4H2O + 0.5eV (2.5μm photon, equivalent to 883°C, in a metastable state) 4H2O + H2O = 5H2O + 0.4eV (3.2μm photon, equivalent to 630°C, in a metastable state) 5H2O + H2O = 6H2O(l) + 0.3eV (4.0μm photon, equivalent to 450°C, stable liquid water molecule clusters) Traditional contact-type condensation heat exchangers absorb all thermal photons locally and transfer them to another medium, not even giving the outside world a chance to observe what’s happening. The condensed water also doesn’t have the opportunity to take priority in accessing these heat photons; otherwise, if it were to hoard them, its temperature would rise by at least 500 degrees!
Reply #22016-05-10
It’s nothing complicated; it’s just a simple reaction equation
Reply #32016-05-11
The text I sent is already very specific

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