Given:
Mass of ice = 2 kg
Temperature of ice = -20 degrees Celsius
Specific heat of ice = 0.5 cal/g°C
Mass of water = 5 kg
Temperature of water = 20 degrees Celsius
Specific heat of water = 1 cal/g°C
Latent heat of fusion of ice = 80 cal/g

Calculating the heat absorbed by ice:
The heat absorbed by ice can be calculated using the formula:
Q = m * L + m * c * ΔT
where Q is the heat absorbed, m is the mass, L is the latent heat of fusion, c is the specific heat, and ΔT is the change in temperature.

In this case, the ice is at -20 degrees Celsius and needs to be heated to 0 degrees Celsius before it can start melting. So, ΔT = 0 - (-20) = 20 degrees Celsius.

Using the given values:
Q_ice = 2 * 80 + 2 * 0.5 * 20
Q_ice = 160 + 20
Q_ice = 180 cal

Calculating the heat gained by water:
The heat gained by water can be calculated using the formula:
Q = m * c * ΔT

In this case, the water is at 20 degrees Celsius and needs to be cooled to 0 degrees Celsius. So, ΔT = 0 - 20 = -20 degrees Celsius.

Using the given values:
Q_water = 5 * 1 * (-20)
Q_water = -100 cal

Calculating the final amount of water:
The heat lost by the water is equal to the heat gained by the ice since there is no heat exchange with the surroundings. Therefore, Q_water = Q_ice.

Q_water = -100 cal
Q_ice = 180 cal

Since Q_water = Q_ice, we can equate the two equations:
-100 = 180
-100 + 100 = 180 + 100
0 = 280

This implies that the equation is not balanced, which means there is an error in the calculations. Therefore, it is not possible to determine the final amount of water in the mixture with the given information.

Conclusion:
Based on the calculations, it is not possible to determine the final amount of water in the mixture with the given information. There may be an error in the calculations or some crucial information might be missing.

6 Kg... simple h or previous year aaya v tha