A lump of ice of 0.1 kg at `-10^(@)C` is put in 0.15 kg of water at `20^(@)C`. How much water and ice will be found in the mixture whenit has reached thermal equilibrium? Specific heat of ice `=0.5kcal kg^(-1).^(@)C^(-1)` and its latent heat of melting `=80` kcal `kg^(-1)`

A piece of ice of mass of 100g and at temperature 0^(@)C is put in 200g of water of 25^(@)C .How much ice will melt as the temperature of the water reaches 0^(@)C ? The specific heat capacity of water = 4200 J kg ^(-1)K^(-1) and the specific latent heat of ice = 3.4 xx 10^(5)J kg^(-1)

A piece of ice of mass of 100g and at temperature 0^(@)C is put in 200g of water of 25^(@) How much ice will melt as the temperature of the water reaches 0^(@)C ? The specific heat capacity of water = 4200 J kg ^(-1)K^(-1) and the latent heat of ice = 3.36 xx 10^(5)J kg^(-1)

A piece of ice of mass of 100g and at temperature 0^(@)C is put in 200g of water of 25^(@) How much ice will melt as the temperature of the water reaches 0^(@)C ? The specific heat capacity of water = 4200 J kg ^(-1)K^(-1) and the latent heat of ice = 3.36 xx 10^(5)J kg^(-1)

A piece of ice of mass of 100g and at temperature 0^(@)C is put in 200g of water of 25^(@) How much ice will melt as the temperature of the water reaches 0^(@)C ? The specific heat capacity of water = 4200 J kg ^(-1)K^(-1) and the latent heat of ice = 3.36 xx 10^(5)J kg^(-1)

A piece of ice of mass 100 g and at temperature 0^@ C is put in 200 g of water of 25^@ C . How much ice will melt as the temperature of the water reaches 0^@ C ? (specific heat capacity of water =4200 J kg^(-1) K^(-1) and latent heat of fusion of ice = 3.4 xx 10^(5) J Kg^(-1) ).

A piece of ice of mass 100 g and at temperature 0^@ C is put in 200 g of water of 25^@ C . How much ice will melt as the temperature of the water reaches 0^@ C ? (specific heat capacity of water =4200 J kg^(-1) K^(-1) and latent heat of fusion of ice = 3.4 xx 10^(5) 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

2kg ice at -20"^(@)C is mixed with 5kg water at 20"^(@)C . Then final amount of water in the mixture will be: [specific heat of ice =0.5 cal//gm "^(@)C , Specific heat of water =1 cal//gm"^(@)C , Latent heat of fusion of ice = 80 cal//gm]

2kg ice at -20"^(@)C is mixed with 5kg water at 20"^(@)C . Then final amount of water in the mixture will be: [specific heat of ice =0.5 cal//gm "^(@)C , Specific heat of water =1 cal//gm"^(@)C , Latent heat of fusion of ice = 80 cal//gm]