If a ball of `80` kg mass hits an ice cube and temperature of ball is `100 .^(@)C`, then how much ice converted into water ? (Specific heat of ball is `0.2 cal g^(-1)`, Latent heat of ice `= 80 cal g^(-1)`)

The entropy change in melting 1g of ice at 0^@C . (Latent heat of ice is 80 cal g^(-1) )

A iron ball of mass 0.2 kg is heated to 100^(@)C and put into a block of ice at 0^(@)C .25 g of ice melts , then specific heat of iron ( in cal kg^(-10) C^(-1) ) is [ Latent heat of fusion of ice = 80 cal g^(-1) ]

1 g of a steam at 100^(@)C melt how much ice at 0^(@)C ? (Length heat of ice = 80 cal//gm and latent heat of steam = 540 cal//gm )

100 gm of water at 20^@C is converted into ice at 0^@C by a refrigerator in 2 hours. What will be the quantity of heat removed per minute? Specific heat of water = 1 cal g^(-1) C^(-1) and latent heat of ice = 80 cal g^(-1)

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.

About 5g of water at 30^(@)C and 5g of ice at -20^(@)C are mixed together in a calorimeter. Calculate final temperature of the mixture. Water equivalent of the calorimeter is negligible. Specific heat of ice =0.5"cal"g^(-1)C^(@)-1) and latent heat of ice =80"cal"g^(-1)

500 gms of water is to be cooled from 3.5^@C to 20^@C . How many grams of ice at -10^@C will be required for this purpose? Specific heat of ice = 0.5 cal g^(-1) ""^@C^(-1) Latent heat of ice = 80 cal g^(-1) ?