PHY- heat(change of state)

Let m be the mass of ice and mass of steam respectively.
Energy required to melt all the ice and increase thetemperature of the 0℃ to 100℃ water
= mLf + mcΔT
= m(3.34 X 105 + 4200 X 100)
= (7.54 X 105)m J
Energy lost to condense all the steam to 100℃ water
= mLv
= (2.26 X 106)m J
Assume there is no heat loss or heat gain from thesurroundings,
The energy lost by steam goes to the energy gain by iceto 100℃ water
When all the ice was turned to 100℃ water, theequilibrium is reached so that there is no net heat exchange between the waterand the steam.
So, energy lost by steam to ice
= (7.54 X 105)m J
And hence, (7.54 X 105)m / (2.26 X 106)m= 1 / 3 of steam is condensed to water of 100℃
So, the ratio of mass of steam to that of water
= (m – m/3) : (m + m/3)
= (2m/3) : (4m/3)
= 1 : 2

good answer, I can learn more from it.