${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{Br} \stackrel{\mathrm{aq}.\mathrm{KOH}}{\to } \mathrm{Product} - \left(\mathrm{I}\right)$
${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{Br} \underset{∆}{\overset{\mathrm{alc}.\mathrm{KOH}}{\to }} \mathrm{Product} - \left(\mathrm{II}\right)$
The mechanism in I and II is respectively

(I) Aqueous medium……${\mathrm{OH}}^{-}$ acts like nucleophilie……… Substitution occurs…Ethanol is formed.
(II) Alcohol medium, high temperature…………. ${\mathrm{RO}}^{-}$ acts like base……. Elimination occurs…..Ethene is formed.