`1kg` ice at `0^(@)C` is mixed with `1kg` of steam at `100^(@)C`. What will be the composition of the system when thermal equilibrium is reached ? Latent heat of fusion of ice `= 3.36xx 10^(6)J kg^(-1)` and latent heat of vaporization of water `= 2.26 xx 10^(6)J kg^(-1)`

Answer

A, C, D

Explanation

Heat obserbed by the ice to raise the temperature `100^(@)C`
`Q_(1) = 1 xx 3.36 xx 10^(5) + 1 xx 4200 xx 100`
` = 3.36 xx 10^(5) + 4.2 xx 10^(5)`
` = (3.36 + 4.2) xx 10^(5)`
` = 7.56 xx 10^(5) = 0.756 xx 10^(6)`
`Q_(2) ` heat released by steam
` = 1 xx 2.26 xx 10^(6)J`
` = 3.26 xx 10^(6)J`
THe extra heat `= Q_(2) - Q_(1)`
` = (2.26 - 0.756 xx 10^(6)`
`= 1.506 xx 10^(6)`
THe arrount of steam condensed
` = (1.506 xx 10^(6))/(2.26 xx 10^(6))`
`= 0.665 kg = 665 gm`
The extra ice` = (1000 - 665) = 335gm`
The amount of ice formed
`= 1000 +335 = 1335g`