Correct Answer - B
Heat lost by water at `20^(@)C=ms_(W)DeltaT`
`H=5xx1xx20=100 kcal`
At first, ice at `(-20^(@)C)` will take heat to change into ice at `0^(@)C`
`H=ms_(ice)DeltaT=(2kg)xx0.5xx20=20` kcal
`:.` After this, heat available `=(100-20)=20kcal`
This heat will now be gained by ice at `0^(@)C`. Let m kg of ice melt.
`:. mxx80=80:.m=1kg`
Out of `2`kg of ice, `1` kg of ice melts into water and `1`kg of ice remains unmelted in container.
`:.` Amount of water in container. `= 5+1= 6kg`