I have been thinking that since the core of the sun maintains its temperature at 15 million degrees Kelvin, then every cubic centimeter of this core is receiving a certain amount of energy to keep it at this temperature. So I was thinking what would happen if we could take a small sphere, say, 0.6 cm in radius (so the area is 1 square cm) and put it in the earth's atmosphere. Would it radiate the same energy that it consumed to get to 15 million degrees in the first place ?

I made some calculations using Stefan-Boltzmann law, here's what I got :

$$ P = Area \times 5.67 \times 10^{-8} \times e \times T^4 $$ So for easier calculations let's assume that it is a black body, and the surrounding temperature is negligible compared to 15 million degrees.

so I got : $$ P = .0001 \times 5.7*10^{-8} \times ({15*10^{6}})^4 = 2.9\times10^{17} \,\mathrm{joules/second} $$

Now, isn't this a huge amount of energy ? I know the body has to be maintained at 15 million degrees to keep radiating energy at this rate, about 0.8 cubic cm of the sun core is about 120 kilograms, so the more important question is for how long will this rate of radiation continue.

I honestly feel that there are mistakes in what I have calculated, so please any correction would be greatly appreciated.

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    $\begingroup$ Worth noting is that while the temperature at the sun's core is quite high, the power density is surprisingly low — only a few hundred watts per cubic meter. This adds up to a lot of power because the core of the sun is still very, very large. $\endgroup$ – rob May 24 '14 at 22:52
  • $\begingroup$ Yes I understand that. But it takes energy to heat the material in the core to that degree, so shouldn't we get the same amount of energy when we let this material cool again ? If using Stefan-Boltzmann equation for calculating that is unclear because we don't know the time it takes for the material to cool, then should we use the specific heat capacity instead ? $\endgroup$ – Abanob Ebrahim May 25 '14 at 8:02
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    $\begingroup$ Yes, heat capacity matters; if your computed power is correct, the next step is to ask how long it would take to cool off. $\endgroup$ – rob May 25 '14 at 14:32

Yes a small marble of material from the center the sun would radiate hugely, and cool off quickly, if it was not at the center of the sun. But since it is at the center, it receives an almost equal radiation from its surroundings and does not cool off.

The marble does not contain all that much energy. It would be something like a nuclear bomb. A bomb creates a momentary nuclear reaction that heats a small amount of material to a temperature around 15 million degrees. It then cools off, transferring that energy to the surrounding country side. See this and this.

In the sun, the marble produces power continuously, but the power is tiny. Volume for volume, the sum produces heat at about the same rate as a compost heap. It produces a lot of heat because it is large. See this Wikipedia article on the Sun.

The reason center of the Sun is so hot it that it has very poor cooling. Energy produced in the center either leaves or raises the temperature of the center. It takes millions of years for energy to reach the surface.

Likewise, the surface of the Sun has poor cooling. The power radiated by the surface of the sun is large because the surface is large.

The power per square meter is that of a black body radiator at 6000 K. See this calculator. It is about $7 MW/m^2$, or $7 W/mm^2$, which is not small. Energy leaves the surface of the Sun only through radiation. This makes the surface heat up until there is enough radiation.

Some computer chips give off more than 7 Watts of waste heat through each square millimeter of surface. With good cooling, they stay below 200C. If space was full of air and you had a giant fan, you could keep the surface of the sun at that temperature.

  • $\begingroup$ Actually the link you provided says that at 6000 K, the power is 70 MW rather than 7 MW. But I also don't understand this number, shouldn't we just divide the sun luminosity by its surface area ? that actually yields ~260 MW per meter cubed. $\endgroup$ – Abanob Ebrahim May 25 '14 at 8:08
  • $\begingroup$ I can't make the numbers work out. If I change to 5778K, that becomes 63 MW/sq m. Luminosity/area = $3.8*10^{26} W$/$6.1 * 10^{24} m^2$ = $62 W/m^2$. There is a missing factor of mega somewhere. Also, mean intensity = $2.0 * 10^7 W/m^2 sr$. Ignoring darkening of the limb, assume the sun radiates over a hemisphere = $2\pi$ sr. That gives $12.6 MW/m^2$. The intensity at 90 deg will be higher, but not by a factor of 5. $\endgroup$ – mmesser314 May 25 '14 at 16:01
  • $\begingroup$ @AbanobEbrahim 260 MW per meter cubed or squared? $\endgroup$ – Jens May 31 '18 at 18:16

The core is under a lot of pressure, so you couldn't remove it from the sun's gravity well without it expanding and cooling to the point fusion stops. Only a small part of the core's energy makes it to the surface.

Each second the sun's core converts 600 million tons of Hydrogen into 595 million tons of helium. The missing 5 million tons is converted to energy. In one second this is equivalent to 1 billion mega-ton hydrogen bombs. This enough energy to fuel America's energy needs for 7 million years.

This energy though takes a long time to escape the core. In fact by the time a photon from the core travels 320 km to a point where the sun is cool enough that electrons bind to atoms again and work it's way to the surface it can take a million years. (link)

The energy managing to escape from the entire surface of the sun is on the order of $3.9*10^{26}J/s$ (based on 5800K and 696,000km radius of sun). Fusion does produce a lot of energy and is the reason you hear it discussed a lot.

  • $\begingroup$ From your statement, one can say that the 5 million tons of mass energy is (roughly) $3.9\times 10^{26} J$, since the surface temperature and the surface area are not changing. $\endgroup$ – LDC3 May 24 '14 at 17:52
  • $\begingroup$ @LDC3 no because that is at the core which is a different temperature and size. $\endgroup$ – user6972 May 24 '14 at 18:28
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    $\begingroup$ The surface doesn't generate energy, only the core. Therefore, the energy dissipated at the surface must come from the core. The only source of energy is from fusion, unless you think some of the energy is coming from the temperature increase due to compression. $\endgroup$ – LDC3 May 24 '14 at 18:39
  • $\begingroup$ @LDC3 Sorry you're right, it's about the same. I meant that some is lost directly from the core in neutrino production in the ppI, ppII and ppIII chains. $\endgroup$ – user6972 May 24 '14 at 21:19
  • $\begingroup$ Did you really mean 210 Kelvinmeter? Or was that a typo? Perhaps you meant to write "210km", though that sounds like too short a distance for the sentence to make sense to me. $\endgroup$ – kasperd Jun 14 '15 at 6:45

Your $15\times10^6$ temperature is at the Sun's core, where H fusion takes place. The radiated energy occurs at the Sun's "surface", at a temperature of only about 5700 K.


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