Why is net radiation heat loss so impossibly high in this everyday case?

Question

If a hot object (temperature $T_h$) is radiating energy to cooler surroundings (temperature $T_c$) engineers express the radiation heat loss as:

$Q = \sigma \varepsilon (T_h^4 - T_c^4) A$

which is directly derived from the black body law. $A$ is the area of the hot object, $\varepsilon$ its emissivity and $\sigma$ the Stefan-Boltzmann constant.

Even though this equation seems very well established and is really just a slightly modified version of the black body radiation equation, today I tried to apply it to a few common everyday cases and the results don’t make sense. They are impossibly high. The most disturbing case is the heat loss (due to radiation) for a human in an environment at room temperature, that I will derive below.

Let’s say a man ($T_h = 37 \deg = 310 \text{ K}$) stands in a room ($T_c = 20 \deg = 293 \text{ K}$). The surface area of a grown-up person has been estimated to be roughly $2 \text { m}^2$. The diffusivity of human skin is also not very well known but 0.9 seems reasonable. And $\sigma$ is $5.67 \times 10^{-8}$ in S.I. units.

Which gives 0.9*(5.67e-8)*2*((310^4)-(293^4)), close to 190 W.

That means the heat lost as radiation by an average dude during a single day is $3600 \times 24 \times 190 \approx 16 \text{ MJ}$. That’s around 4000 kcal: food intake is not even close to simply compensating this huge heat loss (not even speaking about the amount spent as mechanical energy or anything). And that’s in a 20 degrees room! Just imagine if he goes outside on a cold winter day...

What’s wrong here?

Answer 1

If you were a naked person, and your skin temperature was identical to your body temperature, that would probably make some sense (though best case scenario I would only expect ~75% of what you calculated, due to radiation surface area). I'm pretty sure if I was naked in a room that was only 20°C, I would personally start to feel pretty cold.

That said, those two assumptions don't really hold. A quick google search suggests human skin is closer to 33°, so that's already 4 degrees cooler. Some quick math suggests this already reduces the heat transfer by ~25%.

Then you have to consider that your clothes will be warmer than 20°C, so the radiation heat transfer between your body and clothes is actually even lower than the temperature difference between your body and the surrounding room.

I found this paper which gives an equation for calculating the radiation heat transfer from the body (along with other modes of heat transfer). It also mentions that the effective area for body radiation is usually ~75% of the total body surface area. That would be another factor that would reduce the number.

If you want to really dig through some calculations yourself, the linked paper seems to do a really good job organizing all the different modes of heat transfer, and taking into account details like how clothing affects the overall heat transfer.

Answer 2

Radiation is kept back by clothing which both has an insulation because of low thermal conductitity (air kept still) and a lower emissivity. The radiation loss you calculated is true only for a naked body without hair and fur, such a body would indeed get cold at night without radiation from the sun. At daylight, radiation from Sun and clouds (which scatter sun´s radiation) would ease out the problem.