The average person consumes 2000 kcal a day, which is equal to ~100 W. Furthermore, if one uses the Stefan–Boltzmann law to calculate how much someone loses heat due to radiation, it can be seen that it equals

$$Q=\sigma T^4 \varepsilon A$$ $$Q\approx1000\ W$$

Considering a surface area of ~2 m², an emissivity of 0.98 and a temperature of 36.5 °C.

However, this is clearly much greater than the maximum possible heat output of a human body, and that doesn't even consider convection and conduction, which would make heat loss even greater. So what is wrong with this analysis?

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    $\begingroup$ Why do you say 2000 kcal per day rather than using joules? Also, I think you should specify more precisely what you mean by "consume". Do you mean "eat" or "metabolize"? Not all food that is eaten is absorbed, is it? $\endgroup$ Nov 26 '21 at 17:15
  • $\begingroup$ @MatthewChristopherBartsh for most Western diets, full of processed high carb foods and meats it's nearly 100 percent. If you are eating nothing but whole corn, peas and nuts and you don't have good chewing habits it could be much lower. $\endgroup$
    – eps
    Nov 26 '21 at 21:11
  • $\begingroup$ @MatthewChristopherBartsh kcal is the most often used unit when it comes to food consumption. Also, the point of the question was to show that even if all the energy was absorbed and converted into thermal energy it still wouldn't be close to the loss of energy by radiation. $\endgroup$
    – OrdpT
    Nov 29 '21 at 4:10

Your calculation of the radiation power emitted by the human body is correct.

But you forgot, that the human also absorbs radiation from the environment. The walls and all the things in your room probably have a temperature around 20 °C, and therefore emit radiation. The radiation power absorbed by the human body is roughly $$Q_\text{absorbed}=\sigma T_\text{environment}^4 \varepsilon A \approx 840 \text{ W}$$

This absorbed power partially compensates for the emitted power. The net radiation power is $$Q_\text{net} = Q_\text{emitted} - Q_\text{absorbed} \approx 1000 \text{ W} - 840 \text{ W} = 160 \text{ W}$$

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    $\begingroup$ I didn't thought about that! But how did you get the value of 840W? It accounts for a great part of the lost radiation, but the net radiation power still is quite bigger than our energy consumption $\endgroup$
    – OrdpT
    Nov 25 '21 at 22:08
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    $\begingroup$ In my mind, this computation is scholar only. Indeed, this formula only apply to black body radiation and human body and environment are clearly not black body $\endgroup$ Nov 25 '21 at 22:18
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    $\begingroup$ @BaptisteBermond Actually the human body is an almost ideal black body in the infrared range (which is the relevant spectral range for these temperatures). See New World Encyclopedia - Black body - Radiation emitted by a human body $\endgroup$ Nov 25 '21 at 22:59
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    $\begingroup$ Don't forget the surface temperature on most of our bodies is some 20-30 °C when dressed, and goes even less when e.g. some people are running half-naked in winter. They are not losing much heat by radiation; contact with cold air and breathing is probably more significant in such cases. $\endgroup$
    – dominecf
    Nov 26 '21 at 14:56
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    $\begingroup$ My fingertips can get as cold as 20°C when the ambient is about 20°C (so says an IR camera), so I can definitely confirm that you can’t assume that the average exterior body temp is as high as 36.5°C. $\endgroup$ Nov 26 '21 at 21:35

Spherical sheep
This page provides a discussion of heat balance in a spherical sheep, illuminated by Sun and grazing in a meadow. The energy balance equation reads: $$ M + R_+ = R_- + C + H + E,$$ where $M$ is the metabolic energy produced by processing the food, $R_\pm$ is the energy gained and lost via the radiation, $C$ is the energy lost via heat conduction to a surface that the animal is in contact with, $H$ is the energy lost via convection (heating the air around the animal), and $E$ is the energy loss due to the evaporation/sweating.

As we see, there are many paths via which the animal loses its heat. Yet, as we know the warm-blooded animals manage to sustain the temperature above that of the surrounding environment, never coming to equilibrium with it (which should have happened, if the animal were losing heat). Moreover, the animals manage to do useful work - moving, growing, and storing the excess energy supplies as fat. This means that their energy intake is greater than their energy losses, and, under normal conditions, they are not at the risk of disrupting their energy balance.

Net radiative heat gain
As it follows from the calculations in the given link - and here my answer deviates from the other answers given - the animals actually gain heat from the environment, rather than lose it! The reason for that is that, while the animal heat loss can be approximated by the black body radiation, the heat gain is not from the balck body radiation at the ambient temperature! Indeed, the incident radiation comes from many different sources, most of which can be approximated by black bodies, but at temperatures much higher than that of the animal. The main among these is the Sun, and it is the heat absorption of the short-wavelength which is crucial in reversing the radiative heat losses. This point is even more evident, if we think of the cold-blooded animals, such as lizards, which explicitly warm themselves in the Sun in order to be able to be physically active (putting a lizard in a warm dark room does not have the same effect).

Thermal neutral zone
Thermal neutral zone (TNZ) is roughly defined as the region of temperatures where the internal metabolism is sufficient for maintaining the body temperature (without involvement of additional mechanisms, such as shivering when too cold or sweating when too hot). With the standard value of the human body temperature taken to be 34C (or 33C in some sources), the thermal neutral zone stretches a few degrees below and above this value. However, this applies to a naked person - even light clothes significantly reduce the radiative losses, and therefore extend the lower critical temperature of the TNZ to about 18-20C. A naked person in a dark room, protected from external radiation, at 20C would not be able to keep themselves warm, as correctly suggests the calculation by @ThomasFritsch.

enter image description here

  • $\begingroup$ This violates basic thermodynamics. $\endgroup$
    – jamesqf
    Nov 27 '21 at 17:42
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    $\begingroup$ @jamesqf obviously it doesn't, because the environment is not in thermal equilibrium - this is pretty much the point of the answer. It is like a dark object left in the Sun, or the Earth as a whole - which is getting warmer, despite the surrounding vacuum at zero temperature. $\endgroup$ Nov 27 '21 at 19:21

I'll tell you more.

Not only the radiative heat loss of the human body is way above the available heat power, the radiation is clearly not the only mechanism available for losing heat.

Contact heat exchange, air convection (natural and forced), water evaporation from skin and lungs - all these mechanisms work and are important in one situation or another.

What we do in order to stay acceptably warm?

  • we heat our immediate environment when needed.
  • we use clothes in order to block all types of heat exchange. In some cases, a lot of them. Search for "mylar blanket" for an extreme idea of how to reflect most of the radiated infrared back.
  • our body regulates the heat loss by controlling the blood flow to the skin and limbs, lowering the surface temperature when needed. Limbs are in general way below 36.5C, the normal temperature of the hands is below 30C.
  • our breath rythm changes with temperature, too, regulating the evaporative loss of heat
  • if everything else is not enough, our body can elevate the heat production at least twice by forcing muscles to vibrate.
  • 1
    $\begingroup$ One could add that, in a cold environment, the rapid heat loss can have pretty drastic consequences. Appealing again to my Russian background - people falling into cold water (e.g., when washed by a wave from a submarine deck) rarely survive for longer than 10-15 minutes, if they are not promptly recovered. $\endgroup$ Nov 26 '21 at 8:20
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    $\begingroup$ @RogerVadim in this case, the main heat loss mechanism is the water convection and it is faster than all of the above. On the other hand, Russians (when not in military service) like to bath in openings cut in the ice sheet of a frozen river, so there is still some unknown physics to be discovered. $\endgroup$
    – fraxinus
    Nov 26 '21 at 11:43
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    $\begingroup$ This is not really "bathing", but rather "dipping" - one does not stay in water for a long time. An it usually requires extensive preparation for months or years. $\endgroup$ Nov 26 '21 at 11:46
  • $\begingroup$ @Roger Vadim: No, at least if you're Finnish, or one of the American Indian tribes that use a sweat lodge. It just requires sitting in the sauna or sweat lodge for a while - one of the rare circumstances where the body does gain heat from the environment. $\endgroup$
    – jamesqf
    Nov 27 '21 at 17:45
  • $\begingroup$ @jamesqf I don't think that rhis particular Russian tradition is coupled with bania. But what you say about plubging in cold water or snow after bania/sauna is indeed a rather widespread practicel - e.g., in Russia and Germany. $\endgroup$ Nov 28 '21 at 7:53

You're treating the human body as if it were an idealized blackbody object, which leads one to the fallacious conclusion that the human body is 'absorbing' radiation from a cooler environment, in violation of 2LoT in the Clausius Statement sense. Do remember that a warmer object will have higher energy density than a cooler object at all wavelengths, that it is the energy density differential which determines graybody object radiant exitance, and that temperature is a measure of energy density (equal to the fourth root of energy density divided by the radiation constant).

e = T^4 a

a = 4σ/c

e = T^4 4σ/c

T = 4^√e/(4σ/c)

T^4 = e/(4σ/c)

q = ε σ (T^4_h - T^4_c)

∴ q = ε σ ((e_h / (4σ / c)) – (e_c / (4σ / c)))

Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).

W m-2 = W m-2 K-4 * (Δ(J m-3 / (W m-2 K-4 / m sec-1)))

∴ q = (ε c (e_h - e_c)) / 4

Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).

W m-2 = (m sec-1 (ΔJ m-3)) / 4

Take a look at this graphic:


In treating real-world graybody objects as if they were idealized blackbody objects, one is clinging to the long-debunked Prevost Theory of Exchanges and its working principle, the Prevost Principle. Both were chucked on the midden heap of scientific history by none other than James Clerk Maxwell after he read Joule's paper and convinced the scientific community to chuck Caloric Theory (upon which the Prevost Principle is predicated) on the waste pile in favor of the Kinetic Theory Of Heat, which was subsequently superseded by Quantum Thermodynamics.

The Prevost Principle postulates that an object's radiant exitance is predicated only upon the internal state of that object... but that would only work for idealized blackbody objects. A graybody object's radiant exitance isn't solely determined by that object's internal state, as the S-B equation plainly shows.

An idealized blackbody object:

  1. Doesn't actually exist... it's an idealization.

  2. Assumes emission to 0 K

  3. Assumes emissivity = 1 at all times

A graybody object:

  1. Exists

  2. Assumes emission to > 0 K

  3. Assumes emissivity < 1 (and per the definition of emissivity, it is variable with the radiant exitance)

The S-B equation for graybody objects isn’t meant to be used to subtract a fictive ‘cooler to warmerenergy flow from the incorrectly-calculated and thus too high ‘warmer to coolerenergy flow, it’s meant to be used to subtract cooler object energy density (temperature is a measure of energy density, the fourth root of energy density divided by Stefan’s constant) from warmer object energy density. Radiant exitance of the warmer object is predicated upon the energy density gradient.

The problem with the conventional take on radiative energetic exchange is that one must claim that at thermodynamic equilibrium, objects are furiously absorbing and emitting radiation... except that would entail a change in entropy.

That entropy doesn't change at thermodynamic equilibrium means the conventional take on radiative energetic exchange must claim that either entropy does change at thermodynamic equilibrium, or that radiative energetic exchange is an idealized reversible process... neither is the case.

ΔS = ΔQ / T

Only for reversible processes does entropy remain constant. Reversible processes are idealizations. All real-world processes are irreversible.

In reality, entropy doesn't change at thermodynamic equilibrium because radiant exitance falls to zero. Photon chemical potential is zero, Helmholtz Free Energy is zero, no work can be done, no energy can be transferred. The system reaches a quiescent state.

One can use electrical theory to arrive at the same conclusions. Here's a circuit simulator which I created which does so:


You'll note the top two circuits are how the conventional take on radiative energetic exchange is done. It is akin to treating each object as though it were in its own system, unable to interact with the other object (akin to assuming each object emits to 0 K). One must then subtract a wholly-fictive 'cooler to warmer' energy flow from the real (but incorrectly calculated and thus too high because of the assumption of emission to 0 K) 'warmer to cooler' energy flow on the back end to get the equation to balance.

The bottom circuit is the correct way of doing it, it puts both objects into the same system, where they are forced to interact.


Building on the answers from Thomas Fritsch and fraxinus: Alternatively, we could find the ambient temperature at which the human body would not need any heat-control measures such as clothing, unusual exertion, or perspiration. Working the equation in reverse, we can solve for $T_\text{environment} $, given $Q_\text{absorbed} = (1000 \;\text{W} - 100 \;\text{W})$. This yields $T_\text{environment} = 27 °\text{C}$, which seems reasonable to me.

However, as fraxnius points out, the normal temperature of limbs is considerably lower than 36.5°C, so the true answer is more complicated.

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    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ Nov 26 '21 at 16:59
  • $\begingroup$ @JohnRennie This does provide an answer to the question, though I apologize for not being more clear. The question was how the human body can output 1000W in black-body radiation given energy intake of only 100W. I simplified the problem by removing the presence of clothing and active heat management (perspiration, etc.), calculated the required ambient temperature that balances the energy economy, and showed that it comes out to be a reasonable number. $\endgroup$ Nov 26 '21 at 18:52

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