In a container of negligible heat capacity `200` gm ice at `0^(@)`C and 100 gm steam at `100^(@)`C are added to 200 gm of water that has temperature `55^(@)`C. Assume no heat is lost to the surroundings and the pressure in the container is constant of `1.0atm (L_(f)=80 cal//gm,L_(v)=540cal//gm, s_(w)=1 cal//gm^(@)C)`
Amount of the steam left in the system, is equal to

Options

A. `16.7` gm
B. `12.0` gm
C. `8.4` gm
D. 0gm, as there is no steam left

Answer

A

Explanation

Suppose final temperature =100
Then, Heat absorbed by ice `=m(LxxSDeltaT)=200xx180` Cal
Heat absorbed by water `=mSDeltaT=200xx45` Cal
Total heat absorbed `=200xx225` Cal
Suppose 'm' mass of steam condenser then heat returned `=mL_(v)=540xxm=200xx225`
Suppose final temperature=T
then Heat absorbed by ice `=m_(ice)(L_(f)+S.T.)=200(80+T)` Cal
Heat absorbed by water `=m_(water)S(T+55)=200(T-55)` Cal
Heat released by steam `=m_(steam){L_(v)+S(100-)}=100(540+100+T)=100(640-T)` Cal
Heat released by steam= Heat absorbed by ice + water
`Rightarrow 100xx(640-T)=200(80+T)+200(T-55)`
`Rightarrow 640-T=2(80+T)+2(T-55)`
`T=590/5 gt 100^(@)` C Rightarrow All steam has not condensed and hence final temperature it `100^(@)` C
Suppose mass x of steam has condensed. then, Heat released by steam `=540x`
Heat absorbed by ice `=200(80+100)`
Heat absorbed by water `=200(100-55)`
Heat absorbed by steam = Heat absorbed by water and ice
`Rightarrow 540x=200xx225`
`x=250/3`
Water present in final system `=400+x=483.33` gm
Steam present `=100-x=16.67` gm