`5`g of steam at `100^(@)C` is mixed with `10`g of ice at `0^(@)C`. Choose correct alternative /s) :- (Given `s_("water") = 1"cal"//g ^(@)C, L_(F) = 80 cal//g , L_(V) = 540cal//g`)

Options

A. Equilibrium temperature of mixture is `160^(@)C`
B. Equilibrium temperature of mixture is `100^(@)C`
C. At equilibrium ,mixture contains `13(1)/(3)kg` of water
D. At equilibrium ,mixture contains `1(2)/(3) kg` of steam

Answer

B, C, D

Explanation

Required heat
`10` kg ice `(0^(@)C) " "` Available heat
`5` kg steam `(100^(@)C)`
`darr 800 "kcal" " " darr 2700 " Kcal"`
`10` kg water `(0^(@)C) " " 5` kg water `(100 ^(@)C)`
`darr 1000"kcal"`
`10` kg water `(100 ^(@)C)`
So available heat is more than required heat therefore final temperature will be `100^(@)C`.
Mass of steam condensed =`(800+1000)/(540) = (10)/(3)`kg.Total mass of water = `10+ (10)/(3) = (40)/(3) = 13(1)/(3)kg`
Total mass of steam =` 5-(10)/(3) = (5)/(3) = 1(2)/(3) kg`