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Why does temperature not increase by putting more logs into the fire ? It is a very common observation which I have seen but after some research online I found that it is not completely true.

One reason which I think is that if I put more and more logs into the fire , it wouldn’t increase the temperature because the kinetic energy of molecules for all the logs would be same. There is no extra kinetic energy inside the logs or in fire that would increase the total energy of the system. Therefore , I feel hot is because I just see more of that heat present because of the logs in extra amount. Therefore , their temp is same and just that the logs come more near to me. I feel hot. Therefore , the heat which comes on our body depends on the kinetic energy of air particles. So , if their K.E is same.

I am very curious to know if there is a different reason behind this ?

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    $\begingroup$ Thermodynamics usually discusses intrinsic properties amd extrinsic properties. Intrinsic properties do not change if we add more mass having same value for the parameter. (This doesn't answer the question why). $\endgroup$ – AJN Jun 6 at 12:33
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    $\begingroup$ Maybe not the best example to get at the basic principle, since it also has to do with the heat initially pulled away by the logs added, the fuel-vs-air effects, possibly obstructing air flow, and so forth. $\endgroup$ – Pete W Jun 6 at 13:00
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    $\begingroup$ @PeteW Ok. It came in my test. So , is there any simple way to make me understand this phenomenon. $\endgroup$ – Srijan M.T Jun 6 at 13:03
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    $\begingroup$ Why do you think that the temperature doesn't increase when adding more logs? $\endgroup$ – spacetyper Jun 7 at 2:55
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    $\begingroup$ IIRC, the heat from fire is more from breaking down O2 than the hydro-carbon chains of the wood. Perhaps the heat threshold is more due to air injection limitation? $\endgroup$ – chux - Reinstate Monica Jun 7 at 11:51
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Adding more fuel doesn't permit an arbitrarily high combustion temperature because of an unavoidable intrinsic limitation: The reaction has to heat its products. (In the case of wood, this is mostly carbon dioxide and water.)

In the most efficient flame, we might mix the reactants in perfect proportion and eliminate heat losses as best we can, but we can't get around the limitation of heating the products of the combustion reaction. The idealized maximum temperature is called the adiabatic flame temperature and is about 2000°C for wood burning in air. You can estimate this temperature yourself for any reaction if you know the enthalpy of the reaction and the heat capacity of the products.

Other limitations can arise in practice: you can't pack fuel infinitely densely (or, put another way, two identical systems placed side by side aren't twice as hot, as temperature is an intensive variable), the outgoing heat flux increases vastly with increasing temperature, and combustion inefficiencies always exist. But the constraint described above is the ultimate limitation.

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    $\begingroup$ Ok. So , do you mean to say that addition of wood does increase the temperature but to a certain limitation I,e 2000C after which adding more logs doesn’t increase the temperature? $\endgroup$ – Srijan M.T Jun 6 at 17:13
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    $\begingroup$ @Srijan, you're looking for clarification within a minute of the post. Did you read the linked article already? $\endgroup$ – Transistor Jun 6 at 17:15
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    $\begingroup$ @Transistor Oh no. I just read the text. I’ll read the link now. My apologies. $\endgroup$ – Srijan M.T Jun 6 at 17:16
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    $\begingroup$ @SrijanM.T, your question is an interesting one. I had never thought of the problem or the solution but it sounds like a situation where the temperature will rise in an exponential fashion - a bit like charging a capacitor via a resistor - to a maximum value and maybe never actually reaching it. $\endgroup$ – Transistor Jun 6 at 17:22
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    $\begingroup$ Just a general note: if that link (which looked to me just like a link leading onward to more fascinating things about burning...) is critical enough to the answer that it is basically required reading, then I'd suggest to summarize the pertinent information in the answer itself, to be immune against the link going away later. $\endgroup$ – AnoE Jun 7 at 7:17
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I agree with Chemodynamics, and I will try to add a different perspective (or maybe two).

The fire process is a process where you add a fuel (i.e. energy in the system), and that fuel is gradually consumed. The rate at which it is consumed releases the chemical energy. An interesting thing here is that:

  • the log is a three dimensional object
  • the fire will only occur on the external surfaces of the wood.

The above reason is why, if you put small pieces (or shredded) of wood in a fire (because the surface area is larger) the fire seems to grow stronger much faster and the flame is hotter (compared to a log). What happens then is that the flame releases more energy in the unit of time, which heats up the surrounding material (providing the required activation energy) and sustains the combustion.

Perspective No 1:

The energy released by the flame is governed by a square law of the dimensions of the object, while the thermal capacity is governed by the mass (which follows a cube law wrt the dimensions of the object).


At this point its noteworthy that in order to understand fully a combustion process you need to be familiar with concepts like internal energy, enthaphy, specific work due to volumetric changes and others. However, I will try to explain it a more descriptive manner.

The flame of a self sustaining combustion process releases at least enough energy to heat up the surrounding material just enough to be able to ignite (this is called activation energy). A rest of the remaining energy raises a) the temperature of the fuel (and oxidizing agent) even further, and also b) it heats up the environment (which is the useful part).

What you feel as heat when you burn a log is either convective or radiative. The convective heat transfer has to do with the movement of combustion gases (and it increases with the speed of the gases and also the temperature of the gases). The radiative is only depended on the temperature and increases by a 4th order of the temperature. That means that doubling the temperature (in Kelvin) results in 16 times increase in radiative heat losses to the environment.

So assuming 300 K is the ambient temperature, then at 1823 °C (or approx 2100 K), the radiative losses would increase by an order of 74(= 2400).

Perspective No 2:

As the temperature of a flame increases the radiative losses to the environment increase with a power law (4th order). So, despite adding more logs to the pile, once the temperatures reaches a high, the radiative losses are equal to the energy released and there is not enough energy to heat up more the remaining mass in the fire.

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  • $\begingroup$ I like a lot of this answer, but radiative losses do not increase exponentially with temperature. A fourth power is not an exponential. For a large enough argument $x$, any exponential function ($a^x$) grows faster than any finite power ($x^n$). People might be throwing "exponential" around pretty loosely these days, but we should adhere to strict definitions on a engineering Q&A site. $\endgroup$ – WaterMolecule Jun 7 at 21:19
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    $\begingroup$ Again you are right. I will try to find a better wording in the morning. $\endgroup$ – NMech Jun 7 at 22:16
  • $\begingroup$ Persepective No. 1 suggests that we should try a dust explosion $\endgroup$ – Hagen von Eitzen Jun 9 at 11:55
  • $\begingroup$ An injector in car engines more or less does that. $\endgroup$ – NMech Jun 9 at 12:06
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I think a review of the definition of specific heat will help:

Specific heat is the amount of energy required to raise one gram of a pure substance by one degree Centigrade.

$C_P = \frac {Q}{m\Delta T}$,

$Q = Amount$ of $heat$

$m = Mass$

$T = Temperature$ of the $substance$

The specific heat is a constant for each specific material, and the formula can be rearranged as,

$Q = C_Pm\Delta T = m\Delta T$ for $C_P = 1$

Now we can express the relationship between the temperature and the mass pictorially:

enter image description here

In the illustration above, we note that an increase in temperature will increase the heat proportionally, as well as an increase in mass, and the temperature is independent of the mass but the heat.

Hope this helps.

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  • $\begingroup$ According to formula of specific heat capacity, is it that a twice in mass brings 2Q but no need to change $\delta$T ?. Therefore , more Q doesn’t mean an increase in $\deltaT$ unless done specifically an action is done for change in $\delta$ T. $\endgroup$ – Srijan M.T Jun 7 at 4:27
  • $\begingroup$ @Srijan Cp is a constant for the specific substance. If you agree 2*m will produce 2*Q, then Cp = Q/mT' = 2Q/2*mT', where T' represents the change in temperature. $\endgroup$ – r13 Jun 7 at 4:43
  • $\begingroup$ So , 2 Will get cancel right. If there is a change in temperature, then C_p would also change right ? Which you said to be Constant . $\endgroup$ – Srijan M.T Jun 7 at 4:50
  • $\begingroup$ Correct. Cp is a constant by definition. $\endgroup$ – r13 Jun 7 at 4:57
  • $\begingroup$ Yes. So , @r13 . Adding more logs to heat , do you mean to say contribute to more heat. But not that the fire gets hotter or it’s temperature increases. For the 2nd case , do you mean to say that as we increase the temperature, it doesn’t mean there is needed to be a change in mass. Considering 2* for both as you have written. $\endgroup$ – Srijan M.T Jun 7 at 5:08
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Adding more wood does not change the temperature at which wood burns, but it does give you more heat.

Think of it this way: a 15,000 BTU furnace may heat a small apartment, but you'll need a 100,000 BTU furnace to heat a house. They both burn natural gas at the same temperature, the latter just burns more of it at the same time. So the air coming out of both will be the same temperature, but you can get a lot more of it from the larger furnace.

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Let’s say I have a single molecule of CH4 and I want to burn this molecule.

CH4 + 2 O2 -> CO2 + 2 H2O

How much energy is released? Let’s say it releases 1 unit. How much of that unit will be in the form of kinetic energy? Let’s say all of it: 1 unit. How fast is that CO2 moving? How fast are those 2 H2O moving?

Let’s say heat is more or less the random motion of molecules.

So, if I burn 2 molecules of CH4, would the resulting CO2 and H2O molecules be going faster? Nope. There would be double the energy released. You can heat a bigger room (double the size).

I’m just giving a simplified explanation as to why the temperature of the products (the flame) stays the same.

If you want to build a furnace and want to achieve higher temperatures, that’s another story that involves using refractory bricks to isolate (keep the heat in instead of losing it to your surroundings).

//EDIT: Changing Let’s say kinetic energy is more or less the random motion of molecules. to Let’s say heat is more or less the random motion of molecules.

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