In a container of negligible heat capacity, `200 gm` ice at `0^(@)C` and `100gm` steam at `100^(@)C` are added to `200 gm` of water that has temperature `55^(@)C`. Assume no heat is lost to the surrpundings and the pressure in the container is constant `1.0 atm`. (Latent heat of fusion of ice `=80 cal//gm`, Latent heat of vaporization of water `= 540 cal//gm`, Specific heat capacity of ice `= 0.5 cal//gm-K`, Specific heat capacity of water `=1 cal//gm-K)`
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. `0 gm`, as there is no steam left.

Answer

A

Explanation

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`.