Why does a substance with an endothermic heat of solution dissolve?
Dissolution happens in three steps.

Solute-solute attractions must be broken (consumes energy, endotherm), solvent-solvent attractions must be broken (also endotherm), and finally solute-solvent attractions form. This results in a lower energy state and is exotherm.

Dissolution will be endotherm if it takes more energy to break the mentioned attractions, than is released in the last step. You are asking then, why is this reaction spontaneous? In other words, why does it occur? To find out we need chemical thermodynamics. $$\Delta G=\Delta H - T\Delta S$$

$\Delta G$ = Change in Gibb's free energy for a reaction. $\Delta G&lt;0$ = reaction is spontaneous. Otherwise it is NOT spontaneous (ie. will not occur on its own).

$\Delta H$ = Change in enthalpy. $\Delta H &lt;0$ = reaction is exotherm. If larger than 0, the reaction is endotherm. Already now we can see that an endotherm reaction will make $\Delta G$ for the reaction more positive, and thus the reaction is less likely to be spontaneous.

$T$ = temperature in K.

$\Delta S$ = Change in entropy. $\Delta S &gt; 0$ = reaction leads to a state of higher entropy. Entropy describes disorder in a system. Dissolution of a solute in a solvent will always lead to a state of higher disorder, since we go from having all the solute concentrated in a cluster (more ordered) to being spread evenly throughout the solution (more disordered). We can see from the equation that if the reaction leads to higher entropy, then higher temperatures increase likeliness of $\Delta G &lt; 0$, and the reaction will be spontaneous. On the other hand, if we're going to a more ordered state, lower temperatures increase likelihood for a spontaneous reaction.

So how do we apply this to dissolution?
You know that the dissolution reaction is endotherm. That means that $\Delta H &gt; 0$ for this particular reaction. Yet we know that the reaction is spontaneous, since you report that it does indeed occur! This means $\Delta G &lt; 0$. The only way to make that happen is for $\Delta S$ to be positive. So the reaction must go to a more disordered state. Since it's a dissolution, we know that it does. The solute goes from being concentrated in a cluster (ordered) to being spread throughout the solvent (disordered). Pay attention to where temperature is in the equation. It is quite easy to see, that for a dissolution reaction where  $\Delta S &gt; 0$, you can increase the chance of the reaction being spontaneous by increasing temperature!

Bottom line
Your dissolution reaction is endotherm, but because dissolution reactions lead to states of higher entropy, your particular reaction is spontaneous (it occurs) at your specific temperature nonetheless! If it wasn't, heating the system could make it so.