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For example, propane burns in air at 1980 deg C. If an ingot of metal is placed in an insulated container with a hole in the side and a propane flame is directed through that hole, then the metal will heat up. Eventually the metal will approach the temperature of the flame. One might argue that radiative loss of heat out through the hole will equal input of heat when the temperature inside the chamber is equal to the flame. However, the combustion of propane has a certain velocity and the reaction that generates the heat is being 'pumped' into the chamber. Can the temperature inside the chamber rise above the temperature of the combustion reaction since kinetic energy is being continually added to the molecules in the chamber?

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  • $\begingroup$ The heat flow would decrease as the two temperatures approached one another, if somehow the ingot was hotter than the flame, heat would be flowing back toward the flame, which demonstrates that the temperature cannot cross that threshold. If you're compressing the gas continually into the chamber the work done would indeed further raise the temperature but this is a separate energy being added $\endgroup$ – R. Rankin May 18 '18 at 23:29
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Temperature doesn't have an inertia; you can't "gain" on heat transfer induced by a temperature difference to make the lower-temperature system get hotter than the higher-temperature system.

Here, the higher temperature (i.e., the adiabatic flame temperature) is determined by the need to heat the combustion products of propane; this energy use is intrinsic to the burning process and can't be avoided as long as the propane continues to burn in air.

(Note that I'm assuming that the bulk gas velocity is negligible, as would be the case with a typical propane torch. In other words, I'm not considering the stagnation temperature increase from some type of very high-speed jet because it doesn't seem to be the focus of your question. If you wish, however, you can estimate that temperature increase using the link; it appears that the bulk gas velocity would have to be very large indeed for a temperature increase of just a few degrees.)

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  • $\begingroup$ wow, the stagnation temperature is the concept i was imagining. I was surprised to discover that T(stag)/T(static) is a function of the square of the Mach number. but then again the velocity of burning propane is ~50 cm/s =~ 1800 kilometer/h = Mach 1.4 So, yes, the temperature of the ingot would rise above the temp of the burning propane. Thank you for the reference $\endgroup$ – aquagremlin May 20 '18 at 3:06

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