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.

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 .

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 .

Complete the analogy: Latent heat of fusion of ice : 80 cal/g :: steam:

Latent heat of fusion of ice : 80 cal/g :: Steam :…………..

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]

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 -

The water equivalent of a calorimeter is 10 g and it contains 50 g of water at 15^@C . Some amount of ice, initially -10^@C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice = 0.5 cal g^-1^@C^-1, specific heat of water = 1.0 cal g^1^@C^-1 and latent heat of melting of ice =80 cal g^-1 )?