${\mathrm{H}}_2\mathrm{O}$ has higher boiling point than ${\mathrm{H}}_2\mathrm{S}$ because

the latent heat of vaporisation is higher for ${\mathrm{H}}_2\mathrm{O} \mathrm{than} \mathrm{for} {\mathrm{H}}_2\mathrm{S}.$

the bond angle of ${\mathrm{H}}_2\mathrm{O}$ is more than ${\mathrm{H}}_2\mathrm{S}$ and hence ${\mathrm{H}}_2\mathrm{O}$ molecule are more tightly packed

${\mathrm{H}}_2\mathrm{S}$ is a smaller molecule and hence more closely packed

the intermolecular hydrogen bonding in liquid ${\mathrm{H}}_2\mathrm{O}$