# Water vapor condensation on the outside of a water bottle. Endothermic or exothermic and why?

May 6, 2017

Well, it is $\text{exothermic................}$ Why?

#### Explanation:

Chemists are simple folk, and they like to answer problems like this so that the correct solution is OBVIOUS by inspection. So let us try to represent the evaporation of water: i.e. the transition from the liquid phase to the gaseous phase:

${H}_{2} O \left(l\right) \rightarrow {H}_{2} O \left(g\right)$ $\left(i\right)$,

How does this help us? Well, when you put the kettle on to make a cup of tea, CLEARLY you supply energy to boil the water; and to convert SOME of the water to steam. And we can represent this by introducing the symbol, $\Delta$, to represent the heat supplied, i.e.:

${H}_{2} O \left(l\right) + \Delta \rightarrow {H}_{2} O \left(g\right)$ $\left(i i\right)$,

And certainly we can measure this quantity of $\Delta$ in $\text{Joules}$ or even $\text{calories}$. Given the representation of $\text{VAPORIZATION}$, we can reverse the equation to represent $\text{EVAPORATION}$, i.e.

${H}_{2} O \left(g\right) \rightarrow {H}_{2} O \left(l\right) + \Delta$ $\left(i i i\right)$,

We might reasonably intuit that the magnitude of $\Delta$ is IDENTICAL in each case, but in the CONDENSATION reaction, it appears as a product, and in the EVAPORATION it is a virtual reactant.

Given all this (and I apologize for belabouring the point), the reaction as written, $\left(i i i\right)$, is clearly $\text{EXOTHERMIC}$. Capisce?

And in a steam engine we can use this exothermic reaction to do mechanical work. Agreed?