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"Heat rises" or "warm air rises" is a widely used phrase (and widely accepted phenomenon).

Does hot air really rise? Or is it simply displaced by colder (denser) air pulled down by gravity?

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This question has an open bounty worth +50 reputation from NauticalMile ending in 5 days.

The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.

The answers here do not give a satisfactory explanation (in my opinion) on why hot air rises. There should be more mathematical rigor to affirm the previous answers.

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What is the difference? – Mark Eichenlaub Mar 4 '11 at 1:38
    
I suppose the difference is: does thermal motion present in hot air cause it to somehow collectively shimmy its way upwards, displacing the (erstwhile) stationary cold air, or is the true behavior described by the mechanism in his question? – wsc Mar 4 '11 at 1:42
    
I suppose I'm looking for clarification. It strikes me that in the absence of cold(er) air, the only relevant force acting on the hot air would be gravity, pulling it down. Is there another force that I'm missing? – jasonmklug Mar 4 '11 at 1:42
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Yes, WSC... that's about the gist of what I'm wondering. If the cold air were stationary (maybe we assume the cold air is magically unaffected by gravity), would the hot air still rise? – jasonmklug Mar 4 '11 at 1:45
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@Distil: In the absence of other fluid around it, it simply disperses (drifts off into the vacuum) or sit there tightly confines (micro bubbles in some solid). The question only makes sense in the context of a bulk fluid, and then the two cases are one and the same. – dmckee Mar 4 '11 at 2:32
up vote 20 down vote accepted

The mechanism responsible for the rising of hot air is flotation: Hot air is less dense than cold air and hence air pressure will exert an upwards force, in the same way air rises in water. Now if cold air was magically unaffected by gravity, then it would not be able to exert pressure on the hot air and thus it would not rise.

The statement that "heat rises", by the way, is not universally true. Look at water. Here, it is the cold water that is less dense than warm water (at least in the temperature regime of importance to freezing). In winter, when water gets colder, the cold water raises to the top and eventually will freeze, while the water below remains liquid for the moment.

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5  
"heat rises" is generally nonsense! "Heat" is not a stuff, and diffuses. But fighting this wording is like fighting windmills. – Georg Mar 4 '11 at 10:15
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Agreed. The problem is that heat in everyday-language isn't the same as the physical concept of heat. Same with energy, work, order, theory. What can one do... – Lagerbaer Mar 5 '11 at 18:07
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With respect to water, the temperature at which warm water becomes less dense than cold is 4° C. van.physics.illinois.edu/qa/listing.php?id=1736 – Ben Hocking Jun 22 '11 at 18:15

Heat does only 1 thing in a closed system, and that is evenly distribute itself about the system as it reaches thermodynamic equilibrium. I dont think this is what you asking about though. I assume you are talking about hot air (hot being a relative term just meaning it is hotter then the surrounding air). This hot air will be less dense then the surrounding air, and will therefore want to be above the more dense, colder air. If you want to actualy see this, get a beaker of water and add some oil, this is same thing that happens with air (as both cases involve 2 liquids of different densities)

To answer the question exactlty, hot air does rise, and it is also displaced by cold air (though often from the side, not directly above it). And yes, gravity is the reason less dense liquids like to sit on top of more dense liquids

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I have a master's in meteorology, so I can help you out. I apologize if this isn't as professional as some of the other responses, but I'm a little tired right now.

Just take, for example, this equation:

$$F_B = \left( \frac{\rho_0-\rho}{\rho}\right)\approx g\cdot\left( \frac{T-T_0}{T_0}\right)$$

We know that warm air has a lower density than cooler air. So if you want to prove to yourself that warm air has a greater buoyant force than cool air, just plug some numbers in. Just assume that rho-not has a value of 1.25 and that rho has a value of 1.00. That gives you a buoyancy of 0.25. Now, take some cooler air. Increase rho-not to around 1.15. This gives you a buoyancy force of 0.09. So indeed, warm air is more buoyant than cool air and thus experiences a positive buoyancy and rises.

Just keep in mind, though, that this is only valid for parcel theory. Obviously in the real world, there are more things going on than just this equation, but this should at least give you a basic understanding.

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Another way to think about it is to look at how pressure changes with height. If we place a high density box shaped parcel of fluid immediately next to a lower density fluid parcel, the hydrostatic pressure gradient is greater in the former parcel. So, if say the average pressure of the two parcels is the same, the denser one will have higher pressure at the bottom than the lighter one, and lower pressure than the lighter one at the parcel tops. So the denser parcel will tend to push in at the bottom, and be displaced at the top. To a first order approximation the will rotate, trying to put the lighter parcel on top. If the fluid parcels are of identical composition then the warmer one will be lighter. Of course we a temperature regime in water, where the density versus temperature curve runs backwars between roughly 0C to 4C, and ice is lighter still. But in general warmer fluid is lighter.

In anycase the original question is rhetorical. Do we take the mental shortcut and think in terms of bounancy as a lifting force, or try to be more precise and consider the fluids interaction as the cause. In most cases, I'd prefer the former methodology, as that makes it easier to formulate the dynamics.

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I'm going to go over your question bit by bit. Explain some of the language of your question, and then analyze the final answer. My explanations assume prior knowledge of the atomic reality of gases but little else.

First off "heat rises" is a term that should be avoided in a physics discussion. The term "heat" is referring to the transfer of thermal energy from one place to another. It is not a state-quantity. For instance, state quantities are things that are qualities of the matter it self. For instance mass is a state quantity. So is charge. These are the same regardless of other place and time. While "heat" is a description of change, not a description of state. We say a pan on the stove heated up. Or better still a flow of heat from the flame into the pan caused the pan to have a higher temperature. If we said the pan on the stove has heat, that is incorrect, the pan on the stove has thermal energy (a mesure of the mass and temperature of the object), and a temperature.

Reminder: temperature is a mesure of the average kinetic energy of a substance.

Rephrase: Does hot (greater temperature) air rise? Or is it displaced by cold (lower temperature) air?

First: Why does anything fall and rise in a gravitational field? Well it must have a force pushing it up. To change its potential energy (U=mg) a force must act on it.

What is the force that causes a fluid or gas to rise and fall? In all cases it can be described as a pressure.

Pressure is always a relative thing, this is because it isn't pressure that causes things to rise and fall it is a pressure difference or gradient. So what is important is the net pressure, or pressure difference.

First of all this is an important point. If the pressure in a volume is all the same: nothing changes. No air moves (besides individual particles that will move due to brownian motion).

So how can I create a pressure difference to cause one bit of air to rise? To be pushed up?

1) The easiest way is to control how packed the air is, its density. A greater packed group of molecules will have more atoms in a smaller space so it if each molecule is moving at the same speed more collisions occur between the edge of its volume (these changes in momentum cause a force) and it will exert more force: greater pressure.

2) But how do we mesure how fast the particles are going in volume of something? Because if the atoms move faster then there will be greater changes in momentum and more force. Temperature is the mesure of this, the average kinetic energy describes in essence how fast the particles are going.

What does 1) and 2) tell us? Well pressure is controlled by the speed of the particles, and how many of them are in the space. In thermodynamics the equation PV=nRT is used. R is a constant. n is the number of mol (a measure of the number of particles). This says the pressure and volume (V) are related to temperature (speed) and the amount (n).

This says that a hotter volume of the same substance will need to expand to maintain its outward pressure. A colder thing will contract. This is the process of hotter and colder liquids and gases becoming less dense or more dense.

FINAL (Q and A): A: Does hotter air rise? B: Or does cold air displace the hot air causing it to rise?

Well let's test the fist one, hotter implies that it is hotter then something. So if it is hotter then the air around it, the air will expand, the pressure will decrease (since PV is constant) and high pressure, lower density air will push it up: displacing it. Here we see the issue with the question: both A and B are true. If B were not, and cold air wasn't available then there would be no difference in densities, no difference in pressures, and nothing would change. You can not have A without B, and B without A, and mostly this is because a pressure gradient is necessary for changes to occur.

Could you have two gases where the hotter of the two was on the bottom? Yes. A light gas like helium, less dense because its molecules hate each other (personified; sorry), will float all the way out of the earth's atmosphere, leaving hot desert air below.

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The statement "hot air rises" is not in general true, although often used.

Instead,

Less dense air rises

Now usually, locally heated air will expand (because pressure will be similar to the pressure of the surrounding air) according to the universal gas law $PV=nRT$, and less dense air will experience buoyancy from the surrounding more-dense (cooler) air. Hot air will not rise if it's surrounded by hotter air...

Look at the example of a helium balloon, for example. Although the "air" inside the balloon might be colder than the surrounding air, it can still rise - because the gas inside is less dense. And if you created a thin-walled container with low-pressure air (80-20 mixture of nitrogen and oxygen), it could conceivably rise although it's at the same temperature as the surrounding air.

Look also at the air we breathe out: it contains oxygen, carbon dioxide, nitrogen and water. Now carbon dioxide has a higher molecular mass than oxygen, but the addition of water tends to lower the density of the air. So when a politician talks (produces "hot air"), the breath they produce may go up or down. It depends on the temperature of the surrounding air (if the air around him is warmer, for example because he's in a sauna, then the expired air will be cooler than the surrounding air; it may also have lower relative humidity and more carbon dioxide - so it will definitely sink). At sufficiently high relative humidity, in air close to body temperature,(a hot muggy day), it is possible that "hot air sinks".

We can do the math: the composition of air is roughly

in   out  mass  
80%  75%    28  nitrogen
20%  14%    32  oxygen
 0%   6%    18  water
 0%   4%    44  CO2
 1%   1%    40  argon

That makes the average molar mass for inspired air 28.9 g/mol, and 29.8 g/mol for expired air, using the most extreme case of dry air. We can calculate the relative temperatures at which these have the same density:

$$T_1 m_1 = T_2 m_2$$

Using the above numbers, if the temperature of the expired air is 37 °C (330 K), it has the same density as dry atmospheric air with a temperature of 28 °C. This means that when the surrounding air is hotter than 28 C, expired air ("hot air") will sink, even if the relative humidity is zero. It's hard to be a good politician when the airconditioning is broken...

So it's density, not temperature, than matters. Although one often implies the other.

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The Action force is simultaneous with the Reaction force. One can not happen without the other. Archimedes settled that a less dense fluid move on top of a denser fluid (see Buoyancy - Archimedes' principle) .
(and vice-versa: a denser..moves..to bottom..)

The Rayleigh-Taylor instability describes the evolution of the interface between the two layers. The atomic bomb mushroom cap is due to this effect. atomic bomb mushroom cap(quoting Wikipedia):

RT..is an instability of an interface between two fluids of different densities, which occurs when the lighter fluid is pushing the heavier fluid. This is the case with an interstellar cloud and shock system. The equivalent situation occurs when gravity is acting on two fluids of different density — with the dense fluid above a fluid of lesser density — such as water balancing on light oil.

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Boy, does it ever. To contrast the previous answers I will give a mathematical description and a concrete example to bolster the intuitive understanding.

Ideal Gas Law

From thermodynamics we know that pressure, $P$, temperature $T$, and density $\rho$ (or specific volume $v=1/\rho$) are related through an equation of state. For suitable gases (including air at atmospheric conditions) this equation is the ideal gas law:

\begin{equation} \tag{1} \label{igas} P = \rho R T \end{equation}

where $R$ is the specific gas constant, which can be determined by the chemical makeup of the gas under consideration (e.g. $R_{air}=287.058 \:\mathrm{J kg^{−1} K^{−1}} $).

Bouyancy

As already mentioned by Helder Velez, Archemides' Principle informs us that an object immersed in a fluid will experience an upward force equal to the weight of the displaced fluid, where 'up' is the direction of decreasing pressure gradient.1 Mathematically, this may be stated as:

\begin{equation} \tag{2} \label{buoy} \mathbf{F_b} = -\rho V \mathbf{g} \end{equation}

where $\mathbf{g} = -g\mathbf{\hat{k}}$ is the body force vector (usually gravity).

Air Bubble in water

Consider a small air bubble, initially at rest near the bottom of a pool, at thermal equilibrium (same temperature) as the pool water. The buoyancy force acting on the bubble is given by equation \ref{buoy}, and the weight of the bubble is given by $\mathbf{F_g} = m\mathbf{g}$. The subscript $w$ refers to water and the subscript $a$ refers to the air in the bubble. Applying Newton's second law yields:

\begin{align} m_a \mathbf{a} &= \sum \mathbf{F} \\ m_a \mathbf{a} &= \mathbf{F_g} + \mathbf{F_b} \\ m_a \left( a_x \mathbf{\hat{i}} + a_y \mathbf{\hat{j}} + a_z \mathbf{\hat{k}} \right) &= -m_a g\mathbf{\hat{k}} + \rho_w V_b g\mathbf{\hat{k}} \\ m_a a_z &= \rho_w V_a g - m_a g \\ a_z &= g\left(\frac{\rho_w V_a}{m_a} - 1\right) \\ a_z &= g\left(\frac{\rho_w V_a}{\rho_a V_a} - 1\right) \\ a_z &= g\left(\frac{\rho_w}{\rho_a} - 1\right) \\ \end{align}

where I have used $m_a = \rho_a V_a$. Here, it can be seen that the bubble will accelerate upward whenever $\rho_w > \rho_a$. Leveraging the fact the pressure varies linearly with depth in a static fluid, you can prove to yourself that $\rho_w \gg \rho_a$ for bubbles in most pools.

Answer

Parcel of Air

Now consider a similar scenario, where instead of a pool we have a room full of air at uniform temperature $T_\infty$ and the bubble is now a parcel of air which has been heated to a slightly elevated temperature $T_\infty + \Delta T$. I will use the subscripts $c$ for the air in the cool room, and $h$ for the air within the hot parcel.

If we perform a similar analysis to the bubble bubble in the pool, we will go through the same motions as the derivation above, and end with a similar expression for the initial acceleration of the hot parcel:

\begin{equation} a_z = g\left(\frac{\rho_c}{\rho_h} - 1\right) \end{equation}

In this case however, we can use equation \ref{igas} to further simplify the result:

\begin{align} a_z &= g\left(\frac{P/(R_{air} T_\infty)}{P/\left(R_{air} \left[T_\infty + \Delta T\right]\right)} - 1\right) \\ a_z &= g\left(\frac{T_\infty + \Delta T}{T_\infty} - 1\right) \\ \end{align}

I can think of no better mathematical affirmation of the adage Hot air rises than the above equation. Wherever $\Delta T > 0$, $a_z$ will be also. Conversely a cooler parcel will fall: $\Delta T < 0 \rightarrow a_z < 0$.

Cleanup

You might wonder:

Why is it that the bubble in the pool is so straightforward, yet the air parcel rising is not immediately obvious?

Three reasons come to mind:

  1. The bubble is well defined. It has a clear spherical boundary which is more or less maintained during its' ascent. On the other hand, our parcel is not visible, and even if it is spherical initially, it can stretch and morph at the mercy of any local air currents.
  2. The ratio $\rho_w/\rho_a$ is usually much bigger than $\rho_c/\rho_w$, making the motion of the bubble much more pronounced than that of the hot air parcel.
  3. The bubble is subject to heat transfer. Imagine we wrapped our little air parcel in a tiny balloon. Even if its shape is maintained, the air parcel will transfer heat to the surrounding air as it rises, the temperature will drop so that $a_z \rightarrow 0$, and viscous forces will slow it to a halt.

Also note: the magnitude of acceleration is independent of the pressure. Whether we are in a pressure chamber at $10 \:\mathrm{atm}$ or on mount everest at $0.333 \:\mathrm{atm}$ it will always divide through.

Finally, I will point out that, even though the ideal gas law gives us a very elegant expression for acceleration, all other fluids (which I can think of) have equations of state with negative correlations between $T$ and $\rho$, meaning that a fluid parcel with an elevated temperature relative to a quiescent fluid of the same thermodynamic makeup will always have a buoyancy force of greater magnitude than its weight.

1In a stationary reference frame the pressure gradient vector $\nabla P$ of a static fluid is always aligned with the direction of the body force vector $\mathbf{g}$.

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@igael I think I have the correct spelling... it agrees with wikipedia, dictionary.com, and the rest of the answers on the page at least. – NauticalMile 21 hours ago

More correctly, warm air rises from the heat it receives from the Earths surface, even inside. Then it gets too high to receive heat and sinks, while at the same time, the cool air that has just recieved enough energy to rise takes its place.

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