Chemistry - Burns from boiling water and steam

Solution 1:

Let’s consider the following cases:

1. getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water on one’s skin
2. getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ air on one’s skin
3. 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 $40\,\mathrm{°C}$.

1^st 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:

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

2^nd 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.)

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

3^rd 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.)

Solution 2:

After a bit of research, I discovered the following logic:

The boiling point of water is 100 degrees Celsius. At this temperature the water will begin to change from the liquid state to the gas state. In order for this change to take place, additional energy is required. In fact, every gram of liquid water requires 540 calories of heat energy just to convert it to steam. This is called the Heat of Vaporization. During the conversion process, temperature does not increase. It is, therefore, possible to have both liquid water and steam that exist at 100 degrees Celsius. While they both would exist at the same temperature, the steam would have a lot more heat energy due to the additional 540 calories per gram of heat energy that has been absorbed.

This energy release, when made contact with skin, causes a much worse burn than if the same amount of boiling water were to hit your skin where it would decrease in temperature (to your skins temperature) but would not have to go through a phase change. The loss of energy that is released from steam hitting your skin occurs quickly and in a small localized area, therefore causing damage to your cells.