Question

`20 gm` ice at `-10^(@)C` is mixed with `m gm` steam at `100^(@)C`. The minimum value of `m` so that finally all ice and steam converts into water is: `("Use " s_("ice") = 0.5 "cal gm"^(@)C, S_("water") = 1 cal//gm^(@)C, L`) (melting) `= 80 cal// gm` and `L ("vaporization") = 540 cal//gm`)
A. `(85)/(32)gm`
B. `(85)/(64)gm`
C. `(32)/(85)gm`
D. `(64)/(85)gm`

Answer 1

Correct Answer - A
For minimum value of `m`, the final temperature of the mixture must be `0^(@)C`.
`:. 20 xx (1)/(2) xx 10 +20 xx 80 = m 540 +m.1. 100`
`:. M = (1700)/(640) = (85)/(32) gm`.