Applied Thermodynamics in Firefighting
- Page ID
- 418894
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This Exemplar will teach the following concepts from the ACS Examinations Institute General Chemistry ACCM:
A. Most chemical changes are accompanied by a net change of energy of the system.
2. Energy changes can be considered in terms of heat and work.
b. In thermodynamic treatments of chemical systems, the definition of the system of interest versus the surroundings is important.
B. Many chemical reactions require an energy input to be initiated.
C. The type of energy associated with chemical change may be heat, light, or electrical energy.
1. Heat exchange is measured via temperature change.
b. Heat flow is quantitatively obtained from ∆T via molar heat capacity or specific heat and the mass of the substance involved.
What happens when wood burns?
The combustion of wood, like any combustion reaction, releases heat to the surroundings. Just one kilogram of burning wood can release about 2 × 107 Joules of energy.1 This heats the wood and its surroundings: typical wood combustion will have a temperature of around 600-1100℃,2,3 depending on the type of wood and time spent burning. Much of the combustion reaction takes place when wood reaches a temperature at which it can pyrolyze, or decompose into flammable gasses. If you’ve ever sat by a bonfire and seen the fire appear to float above the wood, it’s because the gaseous components produced by pyrolysis are burning.
For wood to ignite and begin combusting, its temperature must rise high enough that pyrolysis takes place and the chemical reaction of combustion starts (around 300 °C).4 Therefore, combustion is a sort of chain reaction: once material starts combusting, it can heat nearby material, and then that material can also start combusting and heat additional material.
Let’s consider the chemistry behind how this works.
Figure \(\PageIndex{1}\). The combustion of wood releases significant heat energy. (Reprinted from image ref 2. 2014, IvanVant via Wikimedia Commons.)
Thermodynamic background
The First Law of Thermodynamics states that energy cannot be created or destroyed. When energy leaves a system, it enters the surroundings, and vice versa — but it is never lost.
If the reaction gives off heat, the surroundings get hotter. If the reaction requires an input of heat, the surroundings cool down. Therefore, they have an inversely proportional relationship. We can represent this using the equation:
ΔEsystem = -ΔEsurroundings
You can think of the change in energy of a system as either the amount of energy leaving the surroundings and entering the system or the amount moving from the surroundings into the system. This transfer is attributed to two sources: kinetic energy change (through heat transfer, written “q”) and potential energy change (through work, written “w”), which gives us:
ΔEsystem = q + w
Now let’s make an assumption. In chemistry, we can often ignore work, since most of the energy change in reactions comes from heat transfer. Usually, work is associated with pressure/volume change work, so in a typical real-world situation (in a room at atmospheric pressure) then there is no pressure change, and there is essentially no change in volume for solutions, so we ignore work in the energy change.
Technically, burning wood creates gas. A balanced reaction of cellulose (the main component of wood)5 combusting would look like this:
6 O2 (g) + C6H10O5 (s) → 6 CO2 (g) + 5 H2O (g)
As we can see, more gas molecules are created than were initially consumed. Thus the volume of the gas in the system does increase. However, throughout the example problems that follow we will ignore this. In fact, substantially less energy is involved in work than heat in these problems (especially as combustion releases a massive amount of energy). Essentially, heat change is so much bigger than the effect of releasing more molecules of gas that work is usually negligible.
Later, we’ll consider the vaporization of water to create steam, which adds gas to the “surroundings.” However, for simplicity we’ll assume the water vapor is freely mixing with the atmosphere and thus the pressure/volume change is negligible. This allows us to consider the work of pressure/volume negligible.
All of that is to say that we can rewrite the initial equation to just:
ΔE = q
How do we calculate the value of q, or ΔE? For any given substance, we can define C, the specific heat, as the amount of energy needed to raise the temperature of 1 gram of the substance by 1℃. In a reaction, we can also measure the mass (m) and the temperature change (ΔT) of the substance or system. This gives us everything we need to know to calculate q, given the useful equation:
q = mCΔT.
Now we can set up another very important equation. We know that ΔEsystem = -ΔEsurroundings and that in chemistry, we can usually make the assumption ΔE = q. So we can say that:
qsystem = -qsurroundings
We can then put these steps together and say:
msys Csys ΔTsys = -msurr Csurr ΔTsurr
So what does this have to do with firefighting? Let’s take a look.
Burning wood heats other wood.
Let’s put the ideas above to work!
If two wooden 2x4s (weighing around 9 pounds each) burn via typical combustion, how much additional wood could be heated from room temperature (22℃) to its flashpoint?
The flashpoint of wood, the lowest possible temperature at which it can burn, is around 300℃.6
Although in reality the combustion reaction of wood will continue to generate heat, we can make a lot of calculations by assuming that all the wood in a system and surroundings reaction comes to thermal equilibrium at the same time. We’ll make this assumption for simplicity purposes throughout the following problems.
Let’s start by defining the system and the surroundings. This can be tricky to do in some real-world situations, and we’ll have to make another assumption here. We’ll define the system as the burning wood, and the surroundings as the additional 2x4s. We’ll make this calculation easier by ignoring the surrounding air and additional materials.
In addition, we’ll assume that the wood burning is not absorbing any of the heat energy it releases; all of the heat released goes to the “surrounding” wood.
Some other information we’ll need here:
-
1lb = 453.592g
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As we noted above, combustion of wood releases 2 × 107 J/kg. Since this energy is being released, we’ll denote it as negative, as is convention in thermodynamic equations.
-
Wood has a specific heat of 1.300 - 2.400 J/g℃ depending on the type.7 Let’s assume an average, 1.85 J/g℃.
Now we can break down the problem:
Let's start by calculating the energy released by combustion.
qsystem = (-2 x 107 J/kg)(1 kg/1000 g)(453.592g /1 lb)(18 lb)
qsystem = -1.63 x 108 J
Now we can set this equal to the right side of the equation qsystem = -msurr Csurr ΔTsurr
-1.63 x 108 J = -msurr(1.85 J/g℃)(300℃ - 22℃)
msurr = 3.18 x 105 g (1 lb/453.592 g) = 699.98 lb of additional wood
That is a lot of wood! If you’ve wondered how a small microwave fire can burn a whole house to the ground, this is your answer. Burning matter releases heat to its surroundings, like other parts of the house, which quickly heat up until they can start burning too. This calculation didn’t even take into account the chain reaction of combustion. The newly heated materials, once combusting, can begin releasing heat to their surroundings, and the fire quickly escalates.
Burning wood also heats the surrounding air.
Now let’s imagine we have half of a 2x4 of wood burning. (This might not seem like a lot, but consider that typically only the outer surfaces of burning wood are actively combusting. This could look like a fairly large fire, since rather than all of any given 2x4 burning up all at once, a small surface of many could burn.)
How hot would the air of a 100 square foot room become with the heat given off from this reaction?
To simplify, let’s again ignore the additional materials in the room and assume all the wood is burning at the same temperature. Once again, the surroundings can be a bit difficult to define in situations like these. For simplicity, we’ll consider all the air in the room as being equally heated by the reaction, and assume that no other materials are absorbing energy. (In reality, obviously nearby air is hotter and further air is cooler, hot air tends to rise, and other materials can absorb energy from the reaction as well.) We can again assume that the air starts at room temperature.
We’ll also ignore the properties of air which change as the temperature increases — that’s beyond our scope here.
Here’s some other information we’ll need:
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Air has a specific heat of 1.005 J/g℃.7
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A 100 sq. ft. room with 12ft ceilings contains 1200 cubic feet
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There are 28 liters per cubic foot at STP -- so we’re dealing with 33,600 liters of air
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1 liter of air weighs 1.293g at STP8
We'll again start by calculating the energy produced by the combustion reaction.
qsystem = (-2 x 107 J/kg)(1 kg/1000 g)(453.592g /1 lb)(9 lb/board)(.5 board)
qsystem = -4.08 x 107 J
Again, we can plug this into our equation qsystem = -msurr Csurr ΔTsurr
-4.08 x 107 J = -msurr Csurr ΔTsurr
-2.04 x 107 J = -(33,600 L)(1.293 g/L)(1.005 J/g℃)(Tfinal - 22℃)
934.45℃ = Tfinal - 22℃
Tfinal = 956.45℃
Consider the implications of this problem. Typical firefighting turnout gear can protect you up to around 590℃.9 Is this a safe room for firefighters to enter? We can conclude that it is definitely not! The temperature in a house fire like this one can quickly rise to several hundred degrees.10 This means that sometimes, the air in the room where the main fire is burning is so hot that the room isn't safe to enter, even in full firefighting gear.
Air has a low specific heat, so it generally rises in temperature quickly. It can also take a lot of air, or a big temperature change, to absorb energy from a reaction.
When Amy was in firefighting academy, a student asked another how houses can still burn in the winter. Let’s investigate:
Imagine the same wood as before is in the same room. How low would the starting temperature of the air need to be in order to stop the spread of combustion?
Again, assume wood has an ignition temperature of 300℃ and consider the surroundings to be the air in the room. The system here must include solid, pyrolyzed, and burnt wood products since we consider that some wood has already burnt and released its energy, but some wood has yet to burn.
Here’s another piece of chemistry you’ll need to solve this problem: in reactions, thermal equilibrium is inevitable. Eventually, the final temperature of the system and surroundings will become equal. We can express this as:
Tfinal,sys = Tfinal,surr
In order to combust, wood must be at or above 300℃. We’ll assume the entire system (solid, pyrolyzed, and burnt wood products) is at the same temperature. This means that the reaction must “settle” at an equilibrium at or below 300℃. Therefore, we know the final temperature of the air must also be at or below 300℃. This is how we’ll solve the equation.
We continue with the same equation as in Example 2.
-1.837 x 108 J = -msurr Csurr (Tfinal - Tinitial)surr
-1.837 x 108 J = -(33,600 L)(1.293g/L)(1.005 J/g℃)(300℃ - Tinitial)
4,207.31℃ = 300℃ - Tinitial
Tinitial = -3,907.31℃
I certainly hope you never experience a winter day this cold! (In fact, I can guarantee you won't -- this temperature is far below absolute zero!11) Houses still burn in cold weather because air is just not that good at absorbing heat.
Figure \(\PageIndex{2}\). A fire can burn even on very cold days. (Reprinted from image ref 1. 2017, Contributor21022018 via Wikimedia Commons.)
What happens when we add water?
In contrast, water is very good at absorbing heat and stopping fire. Water has a specific heat of 4.186 J/g℃.7 This is four times that of air and the highest specific heat capacity of any liquid. (It is much higher than even the specific heat capacity of wood, you’ll notice!) Liquid water is also more dense than air under standard conditions, so the same volume of water has much more mass than air.
To see why water is such a good tool for fighting fire, let’s look at how much energy it absorbs per liter. (As a note, you’ll notice we’ve been using the unit “J” so far in these calculations. J, joules, is a measure of energy. And q, or the heat transfer, is measured in joules.
How much energy per liter do air and water absorb?
Some information we’ll need:
-
The density of water is 1g/cm3
-
1 liter is 1000cm3, so 1L of water weighs 1000g
Therefore, the energy needed to raise 1 liter of water 1℃:
q = mCΔT
q = (1000g)(4.186 J/g℃)(1℃) = 4,186 J (or 4.2 kJ)
Let’s compare this to the energy needed to raise 1 liter of air 1℃. Recall that 1 liter of air weighs 1.293g at STP (it is much less dense than liquid water).
q = mCΔT
q = (1.293g)(1.005 J/g℃)(1℃) = 1.30 J
We can see that it takes about 3,200 times the energy to raise the temperature of a liter of water as compared to a liter of air. This is one of the reasons why water so effectively puts out fire — compared to many substances, it’s able to absorb a lot of heat energy without changing temperature by very much.

Figure \(\PageIndex{3}\). Water is used to absorb the heat energy produced by combustion. (Reprinted from image ref 3. Sylvain Pedneault via Wikimedia Commons.)
Fighting fire
We’ve seen that water “absorbs” the heat from fire much more effectively than air. This is essentially the basic tactic of firefighting: to absorb enough heat from the system that the combustion reaction can no longer take place.
Let’s look at an example of how it works.
Maybe this information inspired you to become a firefighter! Armed with your knowledge of thermodynamics, you pull up to a burning house.
You estimate that about 100 pounds of wood are actively combusting and releasing their energy. You know the wood must be above 300℃ for the combustion reaction to continue. To put out the fire, you’ll need to lower the temperature of the wood below its combustion temperature. How many gallons of water do you need?
Consider for simplicity that this is a closed system. Ignore the nearby air and other materials, and treat the situation as though the water is the only material in the surroundings with capacity to absorb heat. Recall that at thermal equilibrium, the system (wood) and surroundings (water) will reach the same temperature.
At first, it may seem we can solve this problem just the same as the ones above. However, there is another important piece that water contributes to the equation: the conversion of water to steam. This phase change absorbs a lot of energy. In fact, for water at its normal boiling point of 100ºC, the heat of vaporization (ΔHvap) is 2260 J/g. This means that to convert 1g of water at 100ºC to 1g of steam at 100ºC, 2.26 kJ of energy is absorbed.12 That’s a pretty huge amount of energy — and another reason why water is such a great tool for fighting fire!
We’ll also have to consider that liquid water has a different specific heat than gaseous water, around 1.996 J/gºC.13,14
Think of it as a three-step process, if it’s helpful. Energy is absorbed by:
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Liquid water heating (from room temperature, 22ºC, to 100ºC)
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Liquid water vaporizing to steam (at 100ºC)
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Water vapor heating (from 100ºC to the final temperature, 300ºC)
Again setting the energy released by the system (burning wood) equal to the "opposite" of the energy absorbed by the surroundings (water), we can write this as a three-part equation. (Remember that one side of the equation must be negative, as qsys = -qsurr and -qsys = qsurr.)
-ΔEcombustion = mwater Cwater ΔTwater + mwaterΔHvap,water + msteam Csteam ΔTsteam
The mass of water and water vapor will not change through each step, so let’s just rewrite these variables as “m” going forward.
-ΔEcombustion = m(4.186 J/g℃)(100ºC - 22ºC) + m(2260 J/g) + m(1.996 J/g℃)(300ºC - 100ºC)
-ΔEcombustion = m[(4.186 J/g℃)(100ºC - 22ºC) +(2260 J/g) + (1.996 J/g℃)(300ºC - 100ºC)]
-ΔEcombustion = m(326.508 J/g + 2260 J/g + 399.2 J/g)
-ΔEcombustion = m(2985.708 J/g)
-100lb(453.592 g/1lb)(1kg/1000g)(-2 × 107 J/kg) = m(2985.708 J/g)
907,184,000J = m(2985.708 J/g)
m =303,842.17g
A gallon of water weighs around 8 pounds. Now we can finish the calculation:
303,842.17g(1lb/453.592g)(1 gallon/8lb) = 83.73 gallons of water
An average fire engine can carry around 600 gallons of water.4 For bigger fires, most houses have hydrants nearby which engines can hook up to for more water. And additional engines usually arrive on scene quickly!
As we’ve seen, water can absorb the energy from burning wood thousands of times more efficiently than air. Next time you hear about a fire being put out, you can remember what’s actually going on: the burning material and water are moving towards a thermal equilibrium below the material’s combustion temperature. Water absorbs the energy released by the reaction of combustion. Once enough water has been introduced to lower the material below its flash point, the combustion reaction stops. And that’s how fire is fought!
References
- The Energy in Wildfires: The Western United States http://large.stanford.edu/courses/20...per%20kilogram. (accessed 2022 -11 -11)
- Chase. How Hot Does Wood Burn? Examined https://firefighterinsider.com/how-h...burn-examined/. (accessed 2022 -11 -11)
- Palumbo, G.; Parrilli, M.; Ruffo, F. Combustion of wood http://www.whatischemistry.unina.it/en/burn.html. (accessed 2022 -11 -11)
- Know the Difference Between Fire Trucks and Fire Engines? | Fire Blog | The City of Portland, Oregon https://www.portlandoregon.gov/fire/...e%20include%3A. (accessed 2022 -11 -11)
- Cellulose (C6H10O5)n - Structure, Molecular Mass, Properties, Uses https://byjus.com/chemistry/cellulose/. (accessed 2022 -12 -1)
- Fire Engineering Staff. Ignition Temperature of Wood https://www.fireengineering.com/lead...-of-wood/#gref. (accessed 2022 -11 -11)
- Engineering Toolbox. Specific Heat of common Substances https://www.engineeringtoolbox.com/s...ity-d_391.html. (accessed 2022 -11 -11)
- How Much Does a Liter of Air Weigh? https://www.toppr.com/ask/question/a...-will-a-litre/. (accessed 2022 -11 -11)
- Fire-End. What is the Firefighter Turnout Gear Temperature Rating (Heat Rating)? https://www.fire-end.com/blog/2021/0...g-heat-rating/. (accessed 2022 -11 -11)
- International Association Of Fire Chiefs; National Fire Protection Association. Fundamentals of Fire Fighter Skills.; Jones And Bartlett Publishers: Sudbury, Mass., 2009.
- What is Absolute Zero? https://coolcosmos.ipac.caltech.edu/...absolute-zero-. (accessed 2022 -12 -6)
- Heat of Vaporization http://www.kentchemistry.com/links/E...porization.htm. (accessed 2022 -12 -1)
- Water - High Heat Capacity https://bio.libretexts.org/Bookshelv..._Heat_Capacity. (accessed 2022 -11 -11)
- Water - Thermophysical Properties https://www.engineeringtoolbox.com/w...por%3A%201.996. (accessed 2022 -12 -1)
Additional Reading
- Britannica. Dissociation | Chemistry | Britannica. Encyclopædia Britannica; 2019.
- Burning of wood http://virtual.vtt.fi/virtual/innofi...g/burning.html. (accessed 2022 -11 -11)
- Harris, T. How Fire Works https://science.howstuffworks.com/en...sics/fire1.htm. (accessed 2022 -11 -11)
- Moldoveanu, S. C. Wood Pyrolysis - an overview | ScienceDirect Topics https://www.sciencedirect.com/topics...wood-pyrolysis .(accessed 2022 -11 -11)
- Pauld. What is the Temperature of Fire? | News - Target Fire Protection https://www.target-fire.co.uk/resour...usehold%20wood (accessed 2022 -11 -11)
- Petrone, C. BRIDGE Ocean Education https://masweb.vims.edu/bridge/datat...chive0909.html. (accessed 2022 -11 -11)
- Pyrolysis of Wood | Biofuels Academy http://biofuelsacademy.org/index.html%3Fp=658.html. (accessed 2022 -11 -11)
Image References
- Contributor21022018, A fire on the snow. 2017, Wikimedia Commons. https://commons.wikimedia.org/wiki/File:A_fire_on_the_snow._The_North-West_of_Sakhalin.jpg (accessed 2022 -12 -3)
- IvanVant, Wood-burned Snoek. 2014, Wikimedia Commons. https://commons.wikimedia.org/wiki/File:Wood-burned_Snoek.jpg (accessed 2022 -12 -3)
- Sylvain Pedneault, Fire inside an abandoned Convent in Massueville, Quebec, Canada. Wikimedia Commons. https://commons.wikimedia.org/wiki/F...ec,_Canada.jpg (accessed 2022 -12 -5)

