Why don't you get burned by the wood benches in a sauna?

Question

When you go to the sauna you may sit in a room with 90°C+. If it is a "commercial" sauna it will be on for the whole day. How does it come that when you sit on the wood you don't get burned?

I believe this question is different than the "classical" one concerning the "feeling" of heat, which may be explained with a low heat transfer. After a much shorter time other objects seem much "hotter", and the heat transfer is not different (as it's still a room filled with the same air).

My guess would be that the reason is the heat capacity but I cannot really explain it. In my understanding a capacity is the ability to store something (heat, charge, ...). Why should an object be cooler if it can store less heat? Also, cannot this be ignored in this case, as the wood is exposed to the temperature for a very long time?

Answer 1

First of all, I hope you sit on a towel. But even when you touch wood with your bare skin, you don't get burned. This indeed has to do with thermal conductance.

The point is not the heat transfer between the wood and your skin, but rather the heat flowing within the wood. When you touch the surface, your skin and the wood at the very surface equalize their temperature. But because it's only a thin film of wood at the surface, not much heat is transferred. This relatively small amount of heat is quickly transported away from the skin into the body by the high thermal conductance of the human body (many processes play a role here, including blood flow carrying heat away). To further heat up your skin, heat from deeper down in the wood needs to get to the surface, so it can be transferred to your skin. This is the process that is slow whenever a material has low heat conductance, like wood, and allows the skin to transport energy away quicker than it can come from the bulk to the surface, so you don't get burned.

Compare this to touching metal, where the heat stored deep in the bulk of the material can rush to the surface rather quickly, if something cool is touching the surface. Much more heat is transferred and you will burn your hand.

The low heat capacity of a wooden bench certainly also plays a role, simply because if there's little heat stored in the material, it has less energy to heat up your skin with.

Answer 2

Wood is full of air, and air is a terrible conductor of heat. It's not as complicated as it sounds, lighter, i.e. less dense woods, translate heat more poorly than dense ones.

If you look at a cross-section of a piece of wood on the microscopic level, you'll actually see that it's a network of relatively free-floating tubes within a strata of connective resins and polymers, which eventually dry out and allow air to penetrate once removed from the tree. Those tubes are used by the trees to carry things such as nutrients and liquids throughout the plant's various types of stalks, and they are also used to provide structural support. The direction the tubes are going in is the wood's "grain." heat travels down the grain relatively easily, as the tubes are solid pieces from start to end, whereas heat cannot travel very well transversely across the tubes due to the air within and around these tubes being absolutely terrible at conducting heat.

Think of it similarly to the protective ceramic plates used to protect spacecraft upon reentry to the earth's atmosphere. These tiles can reach temperatures of over 2000C, but can be held by an unprotected hand at the same time due to how poorly that heat is conducted through the surface. Skin has water on it, and within it, and water has a very high specific heat, which is a measurement of how many Joules of energy is required in order to heat one gram of material by one degree in the Celsius or Kelvin scales. So our skin has a very high specific heat, meaning it can absorb large quantities of energy while remaining at a fairly constant temperature. Since heat propagates very poorly through materials like the ceramic in question and wood, it's a very simple idea.

There is simply not enough energy being transferred to your skin quickly enough for it to harm you. The medium is incapable of transferring the provided amounts of heat in such a way that it will cause you harm, as the heat that is absorbed by your skin is not replaced by heat residing in other places within the medium due to its incredibly poor conductivity. So, once your skin makes a "cool" spot due to contact, that spot will stay cool, especially considering the fact that water is much more conductive of heat than those other materials, meaning the heat dissipates through your tissues and warms your body rather than burning a single localized spot.

In regards to your query about the wood being exposed for a particularly long time to the same temperature, it is much the same as an object reaching terminal velocity. It is impossible for the object to change when the system it is within does not change. The hotter an object is, the more quickly it will radiate the heat it stores, since "Nature abhors a vacuum." It will eventually reach equilibrium within its system no matter what, so long as the system remains unchanged. If you were to turn the sauna hotter, or cool it down, the temperature of the wood would change gradually, but it will always reach an equilibrium at which point the energy flowing into and out of the wood in the form of heat do not surpass each other.

Answer 3

Analysis

The other answers so far have provided a good intuitive explanation for what's happening in this situation. I want to chime in briefly with the analytical result. It turns out that the theoretical final interface temperature $T$ between two large, uniform, solid objects $\mathrm{A}$ and $\mathrm{B}$ initially at respective surface temperatures $T_\mathrm{A}$ and $T_\mathrm{B}$ is given by* \begin{equation} T=\frac{S_\mathrm{A}T_\mathrm{A}+S_\mathrm{B}T_\mathrm{B}}{S_\mathrm{A}+S_\mathrm{B}}\,, \end{equation} where $S$ is the the thermal effusivity, given by \begin{equation} S=\sqrt{k\rho c_\mathrm{P}}\,, \end{equation} where $k$ is the thermal conductivity (how good the material is at moving heat within itself), $\rho$ is the density (how much of the material is packed into a space), and $c_\mathrm{P}$ is the specific heat capacity (how much heat the material can hold). These three properties are the factors influencing the interface temperature. You can see that the result is basically a weighted average of the initial temperatures using these influences.

Examples

Some representative values for $S$ (in kJ/m^2/K) are 1.1 for human flesh, 0.38 for wood, and 24 for aluminum. With wood starting at 90°C and flesh starting at 35°C, we have a contact temperature of about 49°C. I don't know enough about burn physiology to provide much context to this temperature, but it is almost exactly the maximum recommended value for domestic hot water. The main point is to compare with aluminum at 90°C, for which the contact temperature with flesh works out to 88°C, certainly enough to cause serious harm. Of course, many other factors discussed in other answers will alter these results a bit, but you get the idea.

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*I found a nice derivation and the values in Çengel, but I'm sure there's a good open-source reference out there (thanks to user71659 for the nomenclature tip). Lienhard states (Lienhards state?) the result but don't derive it. Getting to the formula involves some fairly advanced techniques in partial differential equations.

Answer 4

Wood is a poor conductor of heat. The thermal conductivity of wood is relatively low because of the porosity of timber. Thermal conductivity declines as the density of the wood decreases. ... For example, the thermal conductivity of pine in the direction of the grain is 0.22 W/moC, and perpendicular to the grain 0.14 W/moC

Wood across the grain, white pine 0.12
Wood across the grain, balsa 0.055
Wood across the grain, yellow pine, timber 0.147
Wood, oak 0.17
Wool, felt 0.07
Wood wool, slab 0.1 - 0.15 (https://www.engineeringtoolbox.com/thermal-conductivity-d_429.html)