Here ice will absorb heat while hot water will release it.
So if $T$ is the final temperature of the mixture, heat given by water

               $Q_1=mc\Delta T=5\times 1\times \left(30-T\right)$

And heat absorbed by the ice

$Q_2=5\times \left(\frac{1}{2}\right)\left[0-\left(-20\right)\right]+5\times 80+5\times 1\left(T-0\right)$

So, by the principle of calorimetry $Q_1= Q_2,$ i.e.,

                     $150 - 5T = 450 + 5T$

                             $T=- 30 °\mathrm{C}$

Which is impossible as a body cannot be cooled to a temperature below the temperature of the cooling body. The physical reason for this discrepancy is the heat remaining after changing the temperature of ice from -20 to 0^oC with some ice left unmelted and we are taking it for granted that heat is transferred from water at 0^oC to ice at 0^oC so that temperature below 0^oC.
However, as heat cannot flow from one body (water) to the other (ice) at the same temperature (0^oC), the temperature of the system will not fall below 0^oC.