Problem — Fermi estimate: sauna

Problem

In a sauna at $T = 90\,^\circ$C, how much does the average speed of air molecules increase compared to the outside at $T = 20\,^\circ$C? Discuss why a relatively modest increase is already enough to make you sweat.

Solution

The average speed (like all characteristic Maxwell-Boltzmann speeds) scales as the square root of the absolute temperature: $\bar v\propto\sqrt{T}\quad\Longrightarrow\quad \frac{\bar v_\text{sauna}}{\bar v_\text{outside}}=\sqrt{\frac{T_\text{sauna}}{T_\text{outside}}}$ Converting to kelvin: $T_\text{sauna}=90+273=363\ \text{K}\qquad T_\text{outside}=20+273=293\ \text{K}$ Computing the ratio: $\frac{\bar v_\text{sauna}}{\bar v_\text{outside}}=\sqrt{\frac{363}{293}}=\sqrt{1{,}239}=1{,}113$ The average speed increases by only about 11%.

Why it’s enough to make you sweat. In degrees Celsius the jump seems huge (from 20 to 90 °C), but on the absolute scale it is modest, and speed depends on the square root: just +11%. However, what matters for the body is not molecular speed but energy flux. Your normal cooling happens through evaporation of sweat: in the sauna the air is already very hot (often hotter than the skin), so you can no longer give off heat to the air — in fact you receive it. The only remaining route is to sweat heavily, so evaporation carries away the latent heat. The small increase in molecular speed corresponds to a thermal environment that reverses the direction of the exchange, and it is this, not the 11% itself, that triggers sweating. $\ev{\bar v \text{ increases by } +11\%\ \text{approximately}\ \ (\sqrt{363/293}\approx 1{,}113)}$

Connections

Topics: Teoria cinetica dei gas Concepts: Velocità quadratica media · Temperatura Skills: Stime di Fermi