Thermal Radiation

All bodies emit radiation. So why don’t we all shine in the dark?  (right-click and choose “Save Link As…” to download to your computer)


Thermal Radiation and the Human Body

All bodies warmer than absolute zero (-273oC) emit radiation. For $ example an $ average human body (clothed and with an average external temperature of 25 C) emits about 750 W of thermal radiation. The reason we do not notice this loss is that the radiation is invisible, except with a thermographic camera[1] Thermographic camera, [2019-10-09]., and the fact that our surroundings when inside a building at 20 C radiate 700 W back to us. The spectrum of wavelengths and power emitted by a body depends on its temperature. Bodies at temperatures around room temperature emit mainly in the thermal infrared. Temperature above about 800oC are needed to emit visible light. The relationship between the power and wavelength emitted by a body at a given temperature is very well illustrated by the simulation in Ref.[2]Blackbody spectrum PhET simulation, [2019-10-09].. The relationship between the temperature $T$ and the wavelength for which maximum power is emitted, $\lambda_m$, is given by Wien’s Law (Eqn.1):

$\lambda_{m}= \dfrac{(2.90\times10^{-3} \textnormal{ m}\cdot\textnormal{K})}{T} \tag{1} $

Notice that as illustrated by the examples mentioned before the higher the temperature the lower the wavelength of the emitted thermal radiation. The temperature of the sun is 5778 K[3] The Sun, [2019-10-09].. From the formula above we can calculate that the wavelength at maximum power is about 500 nm in the middle of visible spectrum. Clearly the animal vision evolved to be as sensitive as possible to the solar radiation. At the human body temperature (37oC) the maximum is at about 10 $\mu$m so in far infrared. Only some snakes, bats and insects can see this radiation, we have to use a special infrared camera.

The total emitted intensity of radiation (average power emitted per unit area of radiating body) is given by the Stefan-Boltzmann Law[4] Radiation Cooling [2019-10-09]., Eqn.2 :

$I = \epsilon \sigma T^4 \tag{2}$

where $T$ is the surface temperature in degrees Kelvin, $\sigma$ is the Stefan-Boltzmann constant = 5.67 x 108 W/m2K4  and $\epsilon$ is the surface emissivity.

Emissivity is a material property that can vary between 0 and 1, and is strongly dependent on the wavelength. Higher emissivity means that the material will emit or absorb more radiation at this wavelength. As illustrated in our video some materials have high emissivity to visible light and low emissivity in the infrared, and vice versa.  Shiny metallic surfaces have low emissivity both in the visible and infrared. Human skin has an emissivity of ~0.98 in the infrared; the Sun’s surface has an emissivity very close to 1.

An ideal winter paint for a dwelling would have high emissivity in the visible (most energy from the sun comes in visible) and low emissivity in the infrared (the house emits infrared). For the summer we would like to have low emissivity in the visible and high in infrared to save on air conditioning. Unfortunately we do not have commercial paints like this and anyway we would be reluctant to repaint the house twice a year.

A person radiates a power $P_E$.

$P_E = A\epsilon \sigma T^4 = (1.7 \textnormal{ m}^2)(0.98)(5.67\times10^{-8} \textnormal{ W}/ \textnormal{m}^2\textnormal{K}^4)(298 \textnormal{ K})^4 = 750 \textnormal{ W}$

Assuming surface area A=1.7 m2, $\epsilon$=0.98 and the mean surface temperature (clothes and exposed flesh) = 25 C = 298 K

When this person sits in a room at 20 C he or she absorbs a power $P_A$:

$P_A = A\epsilon \sigma T^4 = (1.7 \textnormal{ m}^2)(0.98)(5.67\times10^{-8} \textnormal{ W}/ \textnormal{m}^2\textnormal{K}^4)(293 \textnormal{ K})^4 = 700 \textnormal{ W}$

So the net loss of energy is about 50 W. This is ignoring the effect on clothes but the order of magnitude is consistent with our energy input from food. In reality a dressed person looses about 50% of this heat to radiation and the rest to convection, evaporation and heat exchange due to breathing. The total is about 100 W, or 2000 kcal/day.

What to Notice with a Thermographic Camera

Fig.1. This photograph of the UBC campus was taken with a normal camera on a 24 C day in July.  Some objects appear brighter than others, depending on their colour and the direction of the sunlight.

Fig.2. The same scene as Fig.1, taken at the same time, but with a thermographic camera. Now objects that appear brighter have a higher surface temperature than others, regardless of illumination. For example, the tree at left which is in shadow in Fig.1., is quite warm, and the darker concrete that surrounds it is warmer than the lighter concrete in the foreground, which is reflecting more incident light.

Fig.3. Images of two students taken on a cool day with a normal camera (left) and a thermographic camera (right). The students’ faces are warmer than the environment and show up as bright in the thermal image, regardless of their differing skin colour.

Fig.4. One of the students in Fig.3 photographed on a very hot day. Her face and bare arms appear darker than the concrete on which she is standing, which is at 38 C, and also darker than her clothes. Her blonde hair is getting very hot and is insulating her head from uncomfortable overheating. The different colours in her clothes is not apparent in the thermal image.

A Neat Experiment with Low-E Glass

“Low-E” means low emissivity. Low-E materials don’t radiate thermal energy as efficiently as most materials do for a given temperature; this makes low-E glass ideal for improving the insulation of buildings.

Infrared (IR) thermometers gauge temperature by measuring the intensity of emitted thermal radiation, assuming the emissivity is very high (usually 95%) at the wavelengths they utilize (typically 8-20 μm), which good for most materials. Low-E glass has a coating on one side to reduce the emissivity, thus reducing heat loss due to radiation. Therefore, a comparison of the ambient temperature with the reading on an IR thermometer when pointed at the glass should register a large difference on the low-E side. However, one has to be careful; low-E also means high reflectance, so in an environment of uniform temperature, all you will “see” with the IR thermometer is a reflection of the surroundings and little or no difference will be seen.

Here is how both sides of a sample of low-E glass appear in a room of fairly uniform temperature (Fig.5):

Fig.5. Inside a room, side A, side B and the room itself all appear to be more-or-less the same temperature.

Go outside, where the ground and the sky have radically different radiation temperatures (Fig.6):

Fig.6. In the back garden, the grass and the sky appear to be about 35 C apart in temperature.

Now look at a reflection of the cold sky in both sides of the low-E glass (Fig.7.):

Fig.7. Under a clear sky, the two sides of the low-E glass appear to be 24 C apart in temperature.

Et voilà! A huge difference. The B-side is plainly the low-E side. The A-side IR temperature is dominated by the ambient temperature; the B-side IR temperature is dominated by that of the sky seen in the reflection, and the warm glass itself is not contributing much to the IR radiation.

Thanks to Super Glass of Vancouver[5] Super Glass, [2019-10-09]. for the low-E sample used here.




Updated (CEW) 2019-10-09

Footnotes   [ + ]

1. Thermographic camera, [2019-10-09].
2. Blackbody spectrum PhET simulation, [2019-10-09].
3. The Sun, [2019-10-09].
4. Radiation Cooling [2019-10-09].
5. Super Glass, [2019-10-09].