If 10 g of ice at `0^(@)C` is mixed with 10 g of water at `40^(@)C`. The final mass of water in mixture is (Latent heat of fusion of ice = 80 cel/g, specific heat of water `=1 cal/g""^(@)C`)

50 g of ice at 0^@C is mixed with 50 g of water at 80^@C . The final temperature of the mixture is (latent heat of fusion of ice =80 cal //g , s_(w) = 1 cal //g ^@C)

50 g of ice at 0^@C is mixed with 50 g of water at 80^@C . The final temperature of the mixture is (latent heat of fusion of ice =80 cal //g , s_(w) = 1 cal //g ^@C)

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]

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]

120 g of ice at 0^(@)C is mixed with 100 g of water at 80^(@)C . Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/ g-.^(@)C . The final temperature of the mixture is

120 g of ice at 0^(@)C is mixed with 100 g of water at 80^(@)C . Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/ g-.^(@)C . The final temperature of the mixture is

10 g of ice of 0^(@)C is mixed with 100 g of water at 50^(@)C in a calorimeter. The final temperature of the mixture is [Specific heat of water = 1 cal g^(-1).^(@)C^(-1) , letent of fusion of ice = 80 cal g^(-1) ]