`4 gm` of steam at `100^(@)C` is added to `20 gm` of water at `46^(@)C` in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation `= 540 cal//gm`. Specific heat of water `=1 cal//gm-^(@)C`.

Options

A. 18 gm
B. 20 gm
C. 22 gm
D. 24 gm

Answer

C

Explanation

Heat released by steam inconversion to water at `100^(@)C` is `Q_(1)=mL=4xx540=2160 cal`
Heat required to raise temperature of water from `46^(@)C` to `100^(@)C` is `Q_(2)=mS Deltatheta=20xx1xx54=1080 J`
`Q_(1) gt Q_(2)` and `(Q_(1))/(Q_(2))=2`
Hence all steam is not converted to water only half steam shall be converted to water
`:.` final mass of water =20+2=22 gm