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`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`)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 waterD. At equilibrium ,mixture contains `1(2)/(3) kg` of steam |
Answer» Correct Answer - B::C::D 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` |
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