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

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 beat of fusion of ice = 3.36xx 10^(5)J kg^(-1) and latent head of vaporization of water = 2.26 xx 10^(6)J kg^(-1)

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)

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)

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)

1 kg of ice at 0^(@)C is mixed with 1.5 kg of water at 45^(@)C [latent heat of fusion = 80 cal//gl. Then

1 kg of ice at 0^(@)C is mixed with 1.5 kg of water at 45^(@)C [latent heat of fusion = 80 cal//gl. Then

How should 2kg of water at 60^(@)C be divided so that when one part of it is turned into ice at 0^(@)C it would give out sufficient heat to vaporize the other part? (Specific latent heat of fusionof ice =336xx10^(3)J kg^(-1) and specific latent heat of vaporization of steam =2250xx10^(3)Jkg^(-1) )

What is meant by saying that the latent heat of fusion of ice is 3.34 xx 10^(5)J//kg ?

1kg ice at -10^(@)C is mixed with 1kg water at 100^(@)C . Then final equilirium temperature and mixture content.