Introduction:
When ice at -20 degrees Celsius is added to water at 50 degrees Celsius, heat transfer occurs between the two substances. The ice absorbs heat from the water, causing it to melt, while the water loses heat and cools down. Eventually, the system reaches equilibrium, where both the ice and water are at the same temperature.

Calculation:
To determine the amount of ice and water present at equilibrium, we need to consider the heat gained and lost by each substance.

Heat gained by ice (Q1) = mass of ice (m1) × specific heat capacity of ice (c1) × change in temperature (ΔT1)
Heat lost by water (Q2) = mass of water (m2) × specific heat capacity of water (c2) × change in temperature (ΔT2)

At equilibrium, Q1 = Q2, so we can equate the two equations and solve for the unknowns.

Specific heat capacity:
The specific heat capacity of ice is approximately 2.09 J/g°C.
The specific heat capacity of water is approximately 4.18 J/g°C.

Step 1: Heat gained by ice:
Q1 = m1 × c1 × ΔT1
= 10 g × 2.09 J/g°C × (0 - (-20) °C)
= 10 g × 2.09 J/g°C × 20 °C
= 418 J

Step 2: Heat lost by water:
Q2 = m2 × c2 × ΔT2
= 10 g × 4.18 J/g°C × (0 - 50) °C
= 10 g × 4.18 J/g°C × (-50) °C
= -2090 J

Step 3: Equating heat gained and lost:
Q1 = Q2
418 J = -2090 J

This equation shows that the amount of heat gained by the ice is equal to the amount of heat lost by the water. As a result, 10 grams of ice will melt and mix with the remaining water until thermal equilibrium is reached.

Conclusion:
At equilibrium, all the ice will melt, resulting in a final amount of 20 grams of water. There will be no ice left as it has completely melted. The final temperature at equilibrium will be somewhere between -20 degrees Celsius and 50 degrees Celsius, depending on the specific heat capacities and masses of the ice and water.