aqueous KOH vs. alcoholic KOH

In aqueous KOH state, there are plenty of hydroxide ions formed. The OH- favors the nucleophilic substitution of the halide to form the ROH.
In ethanolic KOH state, instead of forming hydroxide ions, the hydroxide ions will react with the ethanol to form ethanoxide ion: HO- + EtOH = H2O + EtO-. In fact, this ethanoxide is not a good substitute to replace the halide but it rather acts as a strong base to remove the proton from the alkyl group of the haloalkane. In such case, it favors the elimination reaction rather than the substitution reaction.
R-CH2-CH2-Cl = {R-CH-CH2Cl]- = R-CH=CH2 + Cl-
Howeever, it also depends on whether there is any acidic proton in the haloalkane.
I hope you understand what I mean.

Water is a more polar solvent than alcohol. When water is used as the solvent, OH- ion undergoes hydration (i.e OH- ion is surrounded by water molecule due to the attractions between the OH- anion and the positive ends of the more polar water molecules) to a greater extent. This would sharply decrease the basic strength (the strength of accepting a proton) of the OH- ion. Therefore, elimination is less favorable.

When alcohol is used as the solvent, OH- ion undergoes salvation (i.e. OH- ion is surrounded by solvent (alcohol) molecules due to the attractions between the OH- anion and the positive ends of the less polar solvent molecules) to a smaller extent. This would not much decrease the basic strength of the OH- ion. Therefore, elimination is more favorable.