<p>A simple calculation confirms that this value is realistic.</p>
<p>The sun has a power output of 3.8e26 watts.  That energy is generated in the core, which is about 1% of the volume of the sun, but is still about 1.9e25 m³</p>
<p>Therefore there is power generation of about 3.8e26/1.9e25 = 20 watts per cubic metre, averaged over the whole of the core.</p>
<p>The value of 276.5 watts is for the centre of the core, where temperatures and pressures are highest.</p>
<p>You can do a similar calculation for a compost heap, with a radius of (say) one metre, it will have a volume of about 4 m³, and so will generate about 1000 watts.  The compost heap (if a spherical ball in space) would have a surface area of about 12 m², and a luminosity calculator tells me that a black body (a compost heap isn't a black body but whatever) with a area of 12m² and giving of 1000 W, has a temperature of 193K (or -110 F, which is enough to deep freeze the microorganisms that would generate the heat, so this spherical compost heap in a vacuum is not completely realistic)</p>
<p>The conclusion of this is that it is quite possible for the sun to have the same energy generation as a compost heap, yet be incandescent.  And it is possible for your compost heap to have the same energy generation as the sun, and not start glowing.</p>

CC BY-SA 4.0
2023-02-06T19:00:02.430Z
...I read this again. I like that paragraph at the end very much Counterintuitive but woud be a good way to show understanding of the concept. I think your analogy  would make a good physics question for freshman college physics.
Sedumjoy
CC BY-SA 4.0
2022-12-17T23:03:47.953Z
2022-12-17T23:03:47.953Z
James K
<p>Heat is lost through the surface. If you take your <span class="math-container">$1m^3$</span> sphere and your <span class="math-container">$276 W/m^3$</span>, the surface area is approx. <span class="math-container">$4.84 m^2$</span>, which means that you'll have <span class="math-container">$\approx 57 W/m^2$</span>.</p>
<p>If your sphere is <span class="math-container">$10 m^3$</span>, it'll be 2760 W, but the surface area only increase to <span class="math-container">$22.45 m^2$</span>. That works out to <span class="math-container">$\approx123 W/m^2$</span>. If we scale it up to <span class="math-container">$100 m^3$</span>, we have 27.6 kW, but only <span class="math-container">$\approx104 m^2$</span> surface area. That works out to <span class="math-container">$\approx265 W/m^2$</span>.</p>
<p>We could keep going, but I think the pattern is fairly evident: the volume increases faster than the surface area. This can be seen directly from the formulas:</p>
<ul>
<li>Surface area of a sphere: A=<span class="math-container">$4\pi r^2$</span></li>
<li>Volume of a sphere: <span class="math-container">$V=\frac{4}{3} \pi r^3 $</span></li>
</ul>
<p>Volume increases with radius cubed, whilst surface area increases with radius squared.</p>
<p>This means that as your mass grows, the energy output increases far faster than the surface area available for energy loss.</p>

CC BY-SA 4.0
2022-12-19T02:22:14.523Z
The cube-square-law is indeed at the -- well, core of the issue. The sun's surface/volume ratio is 4E-9. As a comparison, a cubic meter has a ratio of 6. The energy produced in a compost heap has lots of surface to escape through. The energy produced in the ginormous core, by contrast, must squeeze through a nanoscopic surface bottleneck to get out.
Peter - Reinstate Monica
CC BY-SA 4.0
2022-12-19T11:05:31.443Z
@Peter-ReinstateMonica: A/V values are meaningless without units.
Eric Duminil
CC BY-SA 4.0
2022-12-19T11:43:59.977Z
@EricDuminil Of course, *in absolute numbers.* Equally obviously, the comparison is unit agnostic: The ratio of the A/V for both bodies is the same, in the order of 1E-9, a dimension-less number.
Peter - Reinstate Monica
CC BY-SA 4.0
2022-12-19T19:07:58.137Z
@Peter-ReinstateMonica How can the ratio of A/V be dimensionless? You are dividing distance^2/distance^3.
WaterMolecule
CC BY-SA 4.0
2022-12-19T20:43:00.823Z
@WaterMolecule A/V is unit dependent and not dimensionless. A1/V1 : A2/V2 is not unit dependent and dimensionless.
Peter - Reinstate Monica
CC BY-SA 4.0
2022-12-19T21:43:50.743Z
@Peter-ReinstateMonica: with your logic, no unit should ever be needed, since you could always compare the values to another measurement with the same unit. Anyway, your comment mentioned absolute values, which don't make sense without unit.
Eric Duminil
CC BY-SA 4.0
2022-12-19T22:00:58.613Z
@EricDuminil True, the numbers were meter based but would be different by just an order of magnitude or so with feet instead. But what's important is the comparison.
Peter - Reinstate Monica
CC BY-SA 4.0
2022-12-19T22:05:03.253Z
And yes, if you have ratios of measurements it's indeed unit independent because they cancel each other out. One length, one duration, one volume etc. is always the same fraction of another length, duration, volume no matter what units you use to measure. In a way you make one length etc. the unit.
Peter - Reinstate Monica
CC BY-SA 4.0
2022-12-20T15:50:47.310Z
Restated, the sun has 1.26e18 times as much surface area, and 1.41e27 times as much volume. The ratio of growth in volume to growth in surface area is the same as the growth in linear dimensions, 1.12e9, no matter what units you use. These are dimensionless scale factors, the units are held in other quantities.
Christopher James Huff
CC BY-SA 4.0
2022-12-17T21:36:57.023Z
2022-12-17T21:36:57.023Z
vidarlo
<p>Yes, the power output of the solar core is about 276.5 watts per cubic metre. However, if we average that power over the whole volume of the Sun it drops to 0.27 watts per cubic metre. (Thanks, ProfRob).</p>
<p>Energy is measured in joules, power is measured in watts. One watt is one joule per second. So (in general) the power tells you how much energy is produced or consumed per second.</p>
<p>A cubic metre of solar core contains a <em>lot</em> of energy, as indicated by its temperature and density, but the rate that it &quot;generates&quot; new energy is rather small. Of course, energy is conserved, so the Sun isn't actually producing new energy, it's  merely converting mass (which is a form of energy) into kinetic energy. Some of that kinetic energy is in the form of photons, and some of it is the kinetic energy of the other fusion reaction products.</p>
<p>The primary fusion reactions operating in the Sun are called the <a href="https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain" rel="noreferrer">proton-proton chain</a> (or p-p chain). Unlike the processes in a hydrogen bomb (which uses deuterium &amp; tritium, not plain hydrogen), the start of the p-p chain is quite slow. When two protons fuse, the resulting diproton is very unstable, and it usually splits apart again. However, in the brief time before the diproton splits there's a tiny probability, on the order of <span class="math-container">$10^{-26}$</span>, that one of the protons in the diproton converts to a neutron, creating a deuteron (a deuterium nucleus). The probability is low because the conversion relies on the weak nuclear force, which is much slower than the strong nuclear force involved in binding the nucleons together.</p>
<p>A typical solar core proton has a half-life of around 10 billion years. The Sun will last a long time because its main reaction process is so slow. That's good news for star longevity, but bad news for anyone who wants to build a fusion reactor running on plain hydrogen.</p>

CC BY-SA 4.0
2022-12-16T20:36:43.353Z
*"its main reaction process is so slow [...] bad news for anyone who wants to build a fusion reactor running on plain hydrogen"* — and now I think I understand why it took 50 years to produce a surplus of energy in terrestrial fusion reactors. I think us non-astronomers see the sun and think, "wow, it's really hot and bright. Fusion = LOTS of energy very fast!" We don't realize it is lots of energy produced slowly from lots of mass, which is also radiated slowly, so it takes a long time to build up to "hot and bright".
Greg Burghardt
CC BY-SA 4.0
2022-12-16T20:49:00.187Z
...Thank you for the wonderful explanation. The compost analogy finally makes good sense now but I have to say the "slowness" of how the sun consumes itself is rather astonishing.
Sedumjoy
CC BY-SA 4.0
2022-12-16T20:51:46.193Z
@Greg Indeed! Comparing P-P and D-T fusion is a bit like comparing a steam engine with an internal combustion engine: they're both based on the Carnot cycle, but there are several important differences.
PM 2Ring
CC BY-SA 4.0
2022-12-16T20:52:46.560Z
Of course, I'm realizing my naive non-astronomer comment is also a naive non-nuclear physicist comment. Each fusion reaction produces a small surplus of energy, but our fusion reactors take a large amount of energy just to keep mass packed together tightly enough to start fusion, and then sustain it. Stars have an advantage here. Gravity is the energy that initiates the reaction and helps sustain it. They are also dense enough that the energy radiates outwards slowly. Let us also not forget that there are a *lot* of cubic meters in the sun.
Greg Burghardt
CC BY-SA 4.0
2022-12-16T20:56:26.927Z
@Sedumjoy Of course, big stars burn a lot faster, and can burn up their hydrogen supply in 10 million years or less. OTOH, red dwarfs can gently burn for over a trillion years.
PM 2Ring
CC BY-SA 4.0
2022-12-16T21:03:15.310Z
@Greg Right. In fact, when the Sun was a proto-star it was much hotter than its present temperature. It took millions of years for it to cool down sufficiently so that it could reach the density & temperature combination needed for sustained fusion. The main thing is temperature, but you also need density: your high KE protons need to bump into each other with high frequency. When the odds of winning the fusion lottery are that low, you need to play often if you to win. :)
PM 2Ring
CC BY-SA 4.0
2022-12-19T09:56:33.997Z
@sedumjoy, `the "slowness" of how the sun consumes itself is rather astonishing` - that's the thing about space; space is big, and suns are, as well. Even if it's only a compost heap per m^2, there are a very great number of compost heaps in the sun. :)
AnoE
CC BY-SA 4.0
2022-12-19T22:19:50.147Z
@GregBurghardt It's exactly the reason why noöne (other than scammers) tries to make a fusion reactor in any way similar to the Sun; it'd be quite useless. The only reason the Sun produces as much power as it does is the sheer mindboggling mass of it. On Earth, we have to look for ways to make things a _lot_ faster - which means both skipping the first incredibly rare step (p+p -> D) and increasing both density and temperature _massively_ .  It's much the same as with fission - we don't replicate the passive generator that is in the Earth, we do something much faster.
Luaan
CC BY-SA 4.0
2022-12-16T20:02:16.807Z
2022-12-17T08:26:27.220Z
2022-12-17T08:26:27.220Z
PM 2Ring
<p>I read an article at abc.net.au saying that the power output of the sun is about 276.5 watts per cubic metre, similar to that of a compost pile.</p>
<p>A compost pile is not incandescent, while the sun is. It's so hot that it glows. So I have a hard time trying to understand how an object that maintains a temperature of 27 million degrees at the core can only have 300 watts of energy per cubic meter at the core. The corona is one million degrees, the photosphere is 11,000 to 6,700 degrees, the chromosphere is 14,000 to 6,700 degrees and the radiative zone closest to the core is 12 million degrees, the radiative zone farthest from the core is 7 million (all temperatures in Fahrenheit).</p>
<p>The total power emitted is 3.8 × 10<sup>26</sup> watts, which makes sense given the mass and temperature. I am sure the article is current. Has anyone seen this before and maybe can offer me a little insight?</p>

CC BY-SA 4.0
2022-12-16T18:49:21.283Z
Your compost pile is small and loses heat to the surroundings fairly quickly. The sun is much larger, and it takes a long time for that 300 Watts to escape, meaning the temperature builds up quite dramatically...
Jon Custer
CC BY-SA 4.0
2022-12-16T19:05:15.327Z
Thank you I see your logic. Let me ponder it but what bothers me is:  The density is 10 time that of lead at the core. So you have a chunck of mass of one cubic  meter that has a temperature of 27 million degrees only have 300 watts of energy? Don't those particles in there have kinetic energy?
Sedumjoy
CC BY-SA 4.0
2022-12-16T19:07:13.240Z
Sure, and so do all the particles around them. Hard to have much heat flow when the temperature gradient is so low.
Jon Custer
CC BY-SA 4.0
2022-12-16T19:09:35.673Z
Ahhh....very good. It is starting to make sense now. Many thanx.
Sedumjoy
CC BY-SA 4.0
2022-12-16T19:14:57.163Z
But, it all adds up to about 62 MW per square meter at the surface of the sun. Remember, the sun is big and has a lot of volume...
Jon Custer
CC BY-SA 4.0
2022-12-16T20:36:48.463Z
A related question on our sister site: https://physics.stackexchange.com/q/370899/123208
PM 2Ring
CC BY-SA 4.0
2022-12-16T20:38:36.620Z
Great link thank you.
Sedumjoy
CC BY-SA 4.0
2022-12-18T14:21:04.530Z
@Michael MacAskill ...thank you for the edits. My prose leaves much to be desired.
Sedumjoy
CC BY-SA 4.0
2022-12-19T07:15:58.600Z
Power, no. Power *density* yes.
hobbs
CC BY-SA 4.0
2022-12-25T01:32:23.430Z
If hay or grain is too damp (> ~12%) it too produces heat as it decomposes, and can actually build up enough heat to ignite and burn the barn down. We're talking about a storage space maybe 20 or 30 feet across.
Arthur Kalliokoski
<the-sun>
Is the power output at the core of the sun about the same as a compost pile (about 300 watts)?
Sedumjoy
# Is the power output at the core of the sun about the same as a compost pile (about 300 watts)?

(Asked by: username_0 on 2022-12-16)

I read an article at abc.net.au saying that the power output of the sun is about 276.5 watts per cubic metre, similar to that of a compost pile.

A compost pile is not incandescent, while the sun is. It's so hot that it glows. So I have a hard time trying to understand how an object that maintains a temperature of 27 million degrees at the core can only have 300 watts of energy per cubic meter at the core. The corona is one million degrees, the photosphere is 11,000 to 6,700 degrees, the chromosphere is 14,000 to 6,700 degrees and the radiative zone closest to the core is 12 million degrees, the radiative zone farthest from the core is 7 million (all temperatures in Fahrenheit).

The total power emitted is 3.8 × 1026 watts, which makes sense given the mass and temperature. I am sure the article is current. Has anyone seen this before and maybe can offer me a little insight?

**Tags:** the-sun

> **Comments:**
> **username_1**: Your compost pile is small and loses heat to the surroundings fairly quickly. The sun is much larger, and it takes a long time for that 300 Watts to escape, meaning the temperature builds up quite dramatically...  
> **username_0**: Thank you I see your logic. Let me ponder it but what bothers me is:  The density is 10 time that of lead at the core. So you have a chunck of mass of one cubic  meter that has a temperature of 27 million degrees only have 300 watts of energy? Don't those particles in there have kinetic energy?  
> **username_1**: Sure, and so do all the particles around them. Hard to have much heat flow when the temperature gradient is so low.  
> **username_0**: Ahhh....very good. It is starting to make sense now. Many thanx.  
> **username_1**: But, it all adds up to about 62 MW per square meter at the surface of the sun. Remember, the sun is big and has a lot of volume...  
> **username_2**: A related question on our sister site: https://physics.stackexchange.com/q/370899/123208  
> **username_0**: Great link thank you.  
> **username_0**: @Michael MacAskill ...thank you for the edits. My prose leaves much to be desired.  
> **username_3**: Power, no. Power *density* yes.  
> **username_4**: If hay or grain is too damp (> ~12%) it too produces heat as it decomposes, and can actually build up enough heat to ignite and burn the barn down. We're talking about a storage space maybe 20 or 30 feet across.  

---

## 3 answers

### Answer by username_2 on 2022-12-16. Score: 38 | Accepted answer

Yes, the power output of the solar core is about 276.5 watts per cubic metre. However, if we average that power over the whole volume of the Sun it drops to 0.27 watts per cubic metre. (Thanks, ProfRob).

Energy is measured in joules, power is measured in watts. One watt is one joule per second. So (in general) the power tells you how much energy is produced or consumed per second.

A cubic metre of solar core contains a *lot* of energy, as indicated by its temperature and density, but the rate that it "generates" new energy is rather small. Of course, energy is conserved, so the Sun isn't actually producing new energy, it's merely converting mass (which is a form of energy) into kinetic energy. Some of that kinetic energy is in the form of photons, and some of it is the kinetic energy of the other fusion reaction products.

The primary fusion reactions operating in the Sun are called the [proton-proton chain](https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain) (or p-p chain). Unlike the processes in a hydrogen bomb (which uses deuterium & tritium, not plain hydrogen), the start of the p-p chain is quite slow. When two protons fuse, the resulting diproton is very unstable, and it usually splits apart again. However, in the brief time before the diproton splits there's a tiny probability, on the order of $10^{-26}$, that one of the protons in the diproton converts to a neutron, creating a deuteron (a deuterium nucleus). The probability is low because the conversion relies on the weak nuclear force, which is much slower than the strong nuclear force involved in binding the nucleons together.

A typical solar core proton has a half-life of around 10 billion years. The Sun will last a long time because its main reaction process is so slow. That's good news for star longevity, but bad news for anyone who wants to build a fusion reactor running on plain hydrogen.

> **Comments:**
> **username_11**: *"its main reaction process is so slow [...] bad news for anyone who wants to build a fusion reactor running on plain hydrogen"* — and now I think I understand why it took 50 years to produce a surplus of energy in terrestrial fusion reactors. I think us non-astronomers see the sun and think, "wow, it's really hot and bright. Fusion = LOTS of energy very fast!" We don't realize it is lots of energy produced slowly from lots of mass, which is also radiated slowly, so it takes a long time to build up to "hot and bright".  
> **username_0**: ...Thank you for the wonderful explanation. The compost analogy finally makes good sense now but I have to say the "slowness" of how the sun consumes itself is rather astonishing.  
> **username_2**: @username_11 Indeed! Comparing P-P and D-T fusion is a bit like comparing a steam engine with an internal combustion engine: they're both based on the Carnot cycle, but there are several important differences.  
> **username_11**: Of course, I'm realizing my naive non-astronomer comment is also a naive non-nuclear physicist comment. Each fusion reaction produces a small surplus of energy, but our fusion reactors take a large amount of energy just to keep mass packed together tightly enough to start fusion, and then sustain it. Stars have an advantage here. Gravity is the energy that initiates the reaction and helps sustain it. They are also dense enough that the energy radiates outwards slowly. Let us also not forget that there are a *lot* of cubic meters in the sun.  
> **username_2**: @username_0 Of course, big stars burn a lot faster, and can burn up their hydrogen supply in 10 million years or less. OTOH, red dwarfs can gently burn for over a trillion years.  
> **username_2**: @username_11 Right. In fact, when the Sun was a proto-star it was much hotter than its present temperature. It took millions of years for it to cool down sufficiently so that it could reach the density & temperature combination needed for sustained fusion. The main thing is temperature, but you also need density: your high KE protons need to bump into each other with high frequency. When the odds of winning the fusion lottery are that low, you need to play often if you to win. :)  
> **username_12**: @username_0, `the "slowness" of how the sun consumes itself is rather astonishing` - that's the thing about space; space is big, and suns are, as well. Even if it's only a compost heap per m^2, there are a very great number of compost heaps in the sun. :)  
> **username_13**: @username_11 It's exactly the reason why noöne (other than scammers) tries to make a fusion reactor in any way similar to the Sun; it'd be quite useless. The only reason the Sun produces as much power as it does is the sheer mindboggling mass of it. On Earth, we have to look for ways to make things a _lot_ faster - which means both skipping the first incredibly rare step (p+p -> D) and increasing both density and temperature _massively_ .  It's much the same as with fission - we don't replicate the passive generator that is in the Earth, we do something much faster.  

---

### Answer by username_5 on 2022-12-17. Score: 6

A simple calculation confirms that this value is realistic.

The sun has a power output of 3.8e26 watts. That energy is generated in the core, which is about 1% of the volume of the sun, but is still about 1.9e25 m³

Therefore there is power generation of about 3.8e26/1.9e25 = 20 watts per cubic metre, averaged over the whole of the core.

The value of 276.5 watts is for the centre of the core, where temperatures and pressures are highest.

You can do a similar calculation for a compost heap, with a radius of (say) one metre, it will have a volume of about 4 m³, and so will generate about 1000 watts. The compost heap (if a spherical ball in space) would have a surface area of about 12 m², and a luminosity calculator tells me that a black body (a compost heap isn't a black body but whatever) with a area of 12m² and giving of 1000 W, has a temperature of 193K (or -110 F, which is enough to deep freeze the microorganisms that would generate the heat, so this spherical compost heap in a vacuum is not completely realistic)

The conclusion of this is that it is quite possible for the sun to have the same energy generation as a compost heap, yet be incandescent. And it is possible for your compost heap to have the same energy generation as the sun, and not start glowing.

> **Comments:**
> **username_0**: ...I read this again. I like that paragraph at the end very much Counterintuitive but woud be a good way to show understanding of the concept. I think your analogy  would make a good physics question for freshman college physics.  

---

### Answer by username_6 on 2022-12-17. Score: 5

Heat is lost through the surface. If you take your $1m^3$ sphere and your $276 W/m^3$, the surface area is approx. $4.84 m^2$, which means that you'll have $\approx 57 W/m^2$.

If your sphere is $10 m^3$, it'll be 2760 W, but the surface area only increase to $22.45 m^2$. That works out to $\approx123 W/m^2$. If we scale it up to $100 m^3$, we have 27.6 kW, but only $\approx104 m^2$ surface area. That works out to $\approx265 W/m^2$.

We could keep going, but I think the pattern is fairly evident: the volume increases faster than the surface area. This can be seen directly from the formulas:

* Surface area of a sphere: A=$4\pi r^2$
* Volume of a sphere: $V=\frac{4}{3} \pi r^3 $

Volume increases with radius cubed, whilst surface area increases with radius squared.

This means that as your mass grows, the energy output increases far faster than the surface area available for energy loss.

> **Comments:**
> **username_7**: The cube-square-law is indeed at the -- well, core of the issue. The sun's surface/volume ratio is 4E-9. As a comparison, a cubic meter has a ratio of 6. The energy produced in a compost heap has lots of surface to escape through. The energy produced in the ginormous core, by contrast, must squeeze through a nanoscopic surface bottleneck to get out.  
> **username_8**: @username_7-ReinstateMonica: A/V values are meaningless without units.  
> **username_7**: @username_8 Of course, *in absolute numbers.* Equally obviously, the comparison is unit agnostic: The ratio of the A/V for both bodies is the same, in the order of 1E-9, a dimension-less number.  
> **username_9**: @username_7-ReinstateMonica How can the ratio of A/V be dimensionless? You are dividing distance^2/distance^3.  
> **username_7**: @username_9 A/V is unit dependent and not dimensionless. A1/V1 : A2/V2 is not unit dependent and dimensionless.  
> **username_8**: @username_7-ReinstateMonica: with your logic, no unit should ever be needed, since you could always compare the values to another measurement with the same unit. Anyway, your comment mentioned absolute values, which don't make sense without unit.  
> **username_7**: @username_8 True, the numbers were meter based but would be different by just an order of magnitude or so with feet instead. But what's important is the comparison.  
> **username_7**: And yes, if you have ratios of measurements it's indeed unit independent because they cancel each other out. One length, one duration, one volume etc. is always the same fraction of another length, duration, volume no matter what units you use to measure. In a way you make one length etc. the unit.  
> **username_10**: Restated, the sun has 1.26e18 times as much surface area, and 1.41e27 times as much volume. The ratio of growth in volume to growth in surface area is the same as the growth in linear dimensions, 1.12e9, no matter what units you use. These are dimensionless scale factors, the units are held in other quantities.  

---

