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)` At the final temperature, mass of the total water present in the system, isA. `472.6 gm`B. `483.3 gm`C. `483.6 gm`D. `500 gm`

In a container of negligible heat capacity `200` gm ice at `0^(@)`C an

1.	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)` At the final temperature, mass of the total water present in the system, isA. `472.6 gm`B. `483.3 gm`C. `483.6 gm`D. `500 gm`
Answer» Correct Answer - B As steam has comparatively large amount of heat to provide in the form at latent heat we check what amount of heat is required by the water and ice to go up to `100^(@)C` that is `(m_(i)L +m_(i)S_(w)DeltaT)+(m_(w).S_(w).DeltaT)` `= [(200 xx 80) +(200 xx1xx 100)] +(200 xx1xx 45)` `= 45,000cal` That is given by m mass of steam, then `m_(s)L = 45,000` `m_(s) = (45,000)/(540) = (500)/(6) = 83.3 gm` therefore `83.3 gm` steam converts into water of `100^(@)C`. Total water `= 200 +200 +83.3 = 483.3 gm` steam left `- 16.7 gm`.