`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]`

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 )

500 gm of ice at – 5^(@)C is mixed with 100 gm of water at 20^(@)C when equilibrium is reached, amount of ice in the mixture is:

What will be the final temperature when 150 g of ice at 0^@C is mixed with 300 g of water at 50^@C . Specific heat of water =1 cal//g//^@C . Latent heat of fusion of ice =80 cal//g .

20g stream at 100°C is passed through 15g ice at -12°C. The mass of water in in calorie metre will be nearly (specific heat of ice =0.5 cal/g°C, specific heat of water=1 cal/g°C latent heat of vaporisation 540 cal/g, latent heat of fusion=80 cal/g and neglect heat capacity of calorimeter )

What amount of ice will remains when 52 g ice is added to 100 g of water at 40^(@)C ? Specific heat of water is 1 cal/g and latent heat of fusion of ice is 80 cal/g.

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)

200 g of ice at –20°C is mixed with 500 g of water 20°C in an insulating vessel. Final mass of water in vessel is (specific heat of ice – 0.5 cal g^(–1°)C^(–1) )

10 gm of ice at – 20^(@)C is added to 10 gm of water at 50^(@)C . Specific heat of water = 1 cal//g–.^(@)C , specific heat of ice = 0.5 cal//gm-.^(@)C . Latent heat of ice = 80 cal/gm. Then resulting temperature is -

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