`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 vaporization `= 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 in conversion to water a `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 Delta theta = 20xx1xx54=1080J`
`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=22gm`.