100 g ice at `0^(@)C ` is mixed with 10 g steam at `100^(@)C ` . Find the final temperature and composition .

Explanation

Heat required to melt ice at `0”^(@)C` = `Q_1 = m_1L_1 = 100 xx 80` cal/g - 8000 cal
Heat required to raise the temperature of water from `0”^(@)C` to `100”^(@)C` = `Q_2` = `m_1s_w Delta`
`100g xx 1 cal g”^(@)C xx 100”^(@)C` = 10,000 cal
Maximum heat ice can absorb from steam to reach boiling point = `Q_1 + Q_2 = 18000 cal`
Heat rejected by steam on complete condensation
`Q_3 = m_sL_v = 10 g xx` 540 cal/g = 5400 cal
Heat rejected by water at 100°C cooled to `0”^(@)C`
`Q_4 = m_sS_w Deltat = 10g xx1 calg”^(@)C xx 100` = 1000 cal
Maximum heat can be supplied to steam to ice at `0”^(@)C = Q_3 + Q_4` = (5400 + 1000)cal = 6400 cal
To melt the ice 8000 cal heat is required, but maximum heat supplied by steam 6400 cal is in sufficient to melt the ice. So resulting temperature of mixture is = `0”^(@)C`
Amount of ice melted `=(“Heat supplied to steam”)/(4)`
`(6400 cal)/(80 cal//g)= 80 g`
Ice remains in the mixture =100g-80 g=20 g
Water present in the  mixture =10 g + 80 g =90 g