If we lit a single fire that was fueled by a substance that burns at 500°F, then around that fire we lit directly against it another fire surrounding the original one. This second fire's fuel also burns at 500°F. Would the original fire increase in temperature? If yes, is there any topic related to this event that I can take to research it further?
Thermodynamics – Does More Fire Create a Hotter Fire?
combustiontemperaturethermodynamics
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Just to add to Vineet's answer. Wien's displacement law only tells you the peak wavelength of a hot object's spectrum. The actual emmission spectra is a broad function and so a hot object will emit both longer and shorter wavelengths.
Remember also that a hot object will emit overall more power, and finally you have to include your eye's sensitivity to various wavelengths.
So a piece of hot metal will appear white because although it's peak emmission will be in the infrared it is still emitting a lot of power at shorter wavlengths and your eye is more sensitive to blue than red.
I'll give brief answers to your questions. If you need more detail, you should ask your questions separately.
What's the difference between heat and work at the atomic level? Isn't heat simply work between particles colliding with different momentum against each other?
Treating a substance semi-classically, one can say that at the atomic level, the atoms have a certain position and momentum. Quantum mechanically, even that's dubious because position and momentum are conjugate variables. With regard to heat and work, these don't exist at the atomic level.
Heat and work are processes, not states. Atoms don't contain heat or work. Neither do individual collections of atoms. Heat and work are measures of quantities transferred amongst objects. Objects don't contain heat or work.
Does an increase of pressure also increases the temperature of the gas?
For an ideal gas being compressed adiabatically, the answer is an emphatic yes. For anything else, the answer is sometimes yes, sometimes no. The answer depends on how much heat is being transferred into or out of the gas and on the nature of the gas. If the gas is right at the triple point (ideal gases don't have a triple point), all that compressing the gas adiabatically is going to do is cause some of the gas to turn into liquid or solid.
Excluding water and other special materials, why does a increase of pressure over a solid rises is melting point?
What your teacher told you is nonsense. Increased pressure does not decrease the molecule's motion. What increasing the pressure does do is to decrease the intermolecular distance.
The reason most substances contract when they freeze is because the bonding forces that make a substance become a crystalline solid hold the atoms/molecules closer together than the intermolecular distance at the same temperature in the liquid phase. Increasing the pressure in these substances decreases the intermolecular distance, thereby making it easier for those intermolecular bonding forces that make a substance a solid to take hold.
Water is different. It expands upon freezing. The structure of ice (ice Ih) is very open thanks to the hydrogen-hydrogen bonds in ice. Because ice expands upon freezing at normal pressures, increasing the pressure reduces the freezing point. Increase the pressure beyond about 100 atmospheres and water/ice starts behaving like most other substances. Increase the pressure beyond 3000 atmospheres and something even weirder happens. Now the freezing point drops markedly with increasing pressure. Increase the pressure beyond that and something even weirder happens: The freezing point increases again, this time very sharply increasing with rising pressure. The freezing point is over 600K at a pressure of 100,000 atmospheres.
If the pressure reduces the motion of the particles, how can the inner core have material with higher temperatures (i.e. particles with higher average kinetic energy)?
What your teacher told you was wrong.
Best Answer
By 500 F I assume you are talking about the ignition temperature of the material being burned when it first ignited, not the temperature of the surface and the flame above the burning material after ignition, which would be considerably higher than 500 F (500 F is around the ignition temperature of wood). Once the fuel, say wood, ignites then the surface temperature and flames will be greater than 500 F.
In that case, I would think that making the fire larger (increasing the circumference of the fire) may increase the temperature of the burning surface and flames and the rate of burning of the fire (heat release) as follows:
The figure below is a simplified representation of a burning surface showing heat and mass transfers (based on the Drysdale fire protection textbook "Introduction to Fire Dynamics".) Part of the heat released in the combustion process is fed back to the burning surface to maintain the combustion process. Part of the heat released is lost to the atmosphere, $\dot Q_{atm}$. (a key to the other terms is given below, if you are interested).
One would expect that most of the heat loss occurs above the flame. But some heat loss also occurs to the side of the flame (i.e., around the perimeter of the burning area). Thinking of the perimeter as the surface of a cylinder surrounding the burning area, the larger the perimeter the lower the surface area to volume ratio. That could favor heat retention in the center of the flaming area and an increase in the temperature of the flames at the center.
A more detailed explanation of the figure is below. In particular I draw your attention to the last two equations. The first gives the heat release rate. The greater that rate the higher the burning temperature, all other things being equal. It is proportional to the mass burning rate.
The last equation gives the mass burning rate. Note that the lower the heat loss rate the greater the mass burning rate. The lower surface to volume ratio of the fire the lower the heat loss rate, all other things being equal.
Hope this helps
Key to Terms in Figure:
$\dot Q_{c}$ = rate at which energy (heat) is released in the fire (kW)
$\dot Q_{F}$ = the heat flux supplied by the flame fed back to the fuel surface ($\frac{kW}{m^2}$)
$\dot Q_{L}$ = heat losses expressed as the heat flux through the fuel surface ($\frac {kW}{m^2}$)
$\dot Q_{atm}$ = heat lost to atmosphere (kW)
$\dot m$ = the rate of burning of the fuel ($\frac{kg}{m^{2}.s}$)
According to Drysdale, "...the rate at which energy is released in a fire ($\dot Q_{c}$) is the most important single factor which characterizes its behavior"
The heat release rate ($\dot Q_{C}$) can be roughly related to the rate of burning $\dot m$ and the heat of combustion of the fuel by the following:
$$\dot Q_{c}= x.\dot m . A_{f}\Delta H_{c} $$
where
$A_f$ is the fuel surface area ($m^2$)
$\Delta H_{c}$ is the heat of combustion of the volatiles ($\frac{kJ}{kg}$) and
$x$ is a factor to account for incomplete combustion (<1.0) which is a function of the mixing of air drawn in from the surrounding atmosphere with the volatiles.
The rate of burning can, in turn, be expressed generally as
$$\dot m=\frac{\dot Q_{F}-\dot Q_{L}}{L_{v}}$$
Where $L_{v}$ is the heat required to convert the fuel into volatiles. For a liquid, that would be the latent heat of vaporization.