Burns from boiling water and steam
Let’s consider the following cases:


getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water on one’s skin
getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ air on one’s skin
getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water vapour on one’s skin


With the slightly irrealistic assumption that all of these liberate all their thermal energy to the skin while cooling down to $40\,\mathrm{°C}$.

1st case

Assuming isobaric conditions (constant pressure), the heat energy liberated from $100\,\mathrm{°C}$ water that is cooled to $40\,\mathrm{°C}$ can be calculated as follows:

$$\Delta Q = \Delta T \cdot C_p(\ce{H2O}) \cdot n(\ce{H2O})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 75.327 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 4.5\,\mathrm{kJ}$$

2nd case

A similar formula holds true. However, now we are considering a gas all the way, not a liquid. (And ‘air’ is technically not a single substance, but we can work with it.)

$$\Delta Q = \Delta T \cdot C_p(\mathrm{air}) \cdot n(\mathrm{air})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 29.19 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 1.7\,\mathrm{kJ}$$

3rd case

Unlike the previous cases, we start off with steam, which will condense to water. Therefore, we need to add the heat of vapourisation to the equation.

$$\Delta Q = \Delta T \cdot C_p(\ce{H2O}) \cdot n(\ce{H2O}) + \Delta H_{\mathrm{vap}} \cdot n(\ce{H2O})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 75.327 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol} + 40.66 \frac{\mathrm{kJ}}{\mathrm{mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 45.2\,\mathrm{kJ}$$

You will note that the third case delivers a lot more heat energy than the first two cases.

(Part of the simplifications I used means that I grossly overestimated the heat transferred by hot air. Which is also why one can put one’s hands under a hair dryer and be fine with it.)