Solution:

1. Heat gained by ice to reach 0°C:
\[
Q_{\text{ice heating}} = 10 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 20°C = 100 \, \text{cal}
\]

2. Heat gained by ice to melt:
\[
Q_{\text{ice melting}} = 10 \, \text{g} \cdot 80 \, \text{cal/g} = 800 \, \text{cal}
\]

Total heat gained by ice:
\[
Q_{\text{total ice}} = 100 \, \text{cal} + 800 \, \text{cal} = 900 \, \text{cal}
\]

3. Heat lost by water:
\[
Q_{\text{water cooling}} = 10 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (50 - T) = 500 - 10T \, \text{cal}
\]

4. Set heat gained equal to heat lost:
\[
500 - 10T = 900 \Rightarrow -10T = 400 \Rightarrow T = -40°C
\]

Since this temperature is below 0°C, all the ice will not melt.

5. Equilibrium Calculation:
Let \( x \) be the mass of ice that melts:
\[
500 = 70x + 100 \Rightarrow 400 = 70x \Rightarrow x \approx 5.71 \, \text{g}
\]

Result:
The final temperature of the system will be **0°C**.