Burns from boiling water and steam Why is a burn to the skin caused by steam more serious than a burn caused by the same amount of boiling water at the same temperature?


The temperature is the same, which implies that the kinetic energy of the particles in both the steam and the boiling water are the same; and yet the steam is more dangerous. Why so?
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.)