`5` kg of steam at `100^(@)C` is mixed with `10` kg of ice at `0^(@)C`. Choose correct alternative//s (`Given s_("water") = 1 "cal"//g^(@)C, L_(F) = 80 "cal"//g L_(v)= 540 "cal"//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)g` of water
D. At equilibrium ,mixture contains `1(2)/(3) g` of steam

Answer

B, C, D

Explanation

Required heat `" "` Available heat
`10`g ice (`0^(@)C`) `" "` `5` of steam (`100^(@)C`)
`darr800"cal"` `" "` `darr2700"cal"`
`10` g of water (`0^(@)C`) `" "` `5` g water(`100^(@)C`)
`darr1000"cal"`
`10` g water (`100^(@)`)
So available heat is more than required heat therefore final temperature will be `100^(@)C`.
Mass of vapour condensed
=`(800+1000)/(540) = (10)/(3)g`
Total mass of water
=`10 + (10)/(3) = (40)/(3) = 13(1)/(3)`g
Total mass of steam
= ` 5 - (10)/(3) = (5)/(3) = 1(2)/(3)`g