Stefan–Boltzmann law applied to the human body

Question

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?

Answer 1

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 $$Q_\text{net} = Q_\text{emitted} - Q_\text{absorbed} \approx 1000 \text{ W} - 840 \text{ W} = 160 \text{ W}$$

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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:

https://i.sstatic.net/Qe7Vr.gif

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 warmer’ energy flow from the incorrectly-calculated and thus too high ‘warmer to cooler’ energy 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:

https://tinyurl.com/yzo8hak9

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.