Correct Answer - B
Maximum heat that can be released by water
`=5xx1xx(20-0)=100kcal`
Heat required to change ice from `-20^@C` to `0^@C`
`=2xx0.5[0-(-20)]=20kcal`
Heat required to melt ice `=2xx80=160cal`
`100kcal lt (20+160)kcal`,
whole ice is not melted.
Equilibrium temperature `=0^@C`
Amount of ice melted `=(100-20)/(80)=1kg`
Amount of water `=5+1=6kg`