How do the products differ when ethyl bromide reacts separately with aq. KOH and alc. KOH?

This can be explain in the following ways:

When alkyl chloride react with aqueous KOH it forms alcohol. it is subsitution reaction.

C₂H₅Cl +KOH(aq.) ---> C₂H₅OH +KCl

When alkyl chloride react with alcoholic KOH it forms alkene. In this case elimination reaction take place.

C₂H₅Cl +KOH(alc.) ---> H­₂C=CH₂ +KCl +H₂O