As water equivalent of pitcher is 0.5 kg, i..e., pitcher is equivalent to 0.5 kg of water, heat to be extracted from the system of water and pitcher for decreasing its temperature from 30 to 28^oC is

       ${\text{Q}}_1=\left(\text{m}+\text{M}\right)\text{c}\Delta \text{T}$

           $=\left(\text{9.5}+\text{0.5}\right)\text{kg }\left(1k\; cal/kg{\text{C}}^{\circ }\right){\left(30-28\right)}^{\circ }\text{C}$

           $=20kcal$

And heat extracted from the pitcher through evaporation in t minutes

${\text{Q}}_2=\text{mL}=\left[\frac{\text{dm}}{\text{dt}}\times \text{t}\right]\text{L}=\left[\frac{1\text{g}}{\text{min}}\times \text{t}\right]580\frac{\text{cal}}{\text{g}}$

               $=580\times \text{t cal}$

According to given problem ${\text{Q}}_2={\text{Q}}_1$,  i.e., $580\times \text{t}=20\times 10^3$

                                          t = 34.5 min