A 10 g ice cube is dropped into 45 g of water kept in a glass. If the water was initially at a temperature of `28^(@)C` and the temperature of ice`-15^(@)C`, find the final temperature (in `^(@)C`) of water.
(Specific heat of ice `=0.5" "cal//g-""^(@)C" and "L=80" "cal//g)`

Find the heat required to convert 20 g of iceat 0^(@)C into water at the same temperature. ( Specific latent heat of fusion of ice =80 cal//g )

A colorimeter contains 400 g of water at a temperature of 5^(@)C . Then, 200 g of water at a temperature of +10^(@)C and 400 g of ice at a temperature of -60^(@)C are added. What is the final temperature of the contents of calorimeter? Specific heat capacity of water -1000 cal //kg//K Specific latent heat of fusion of ice =80xx1000 cal //kg Relative specific heat of ice =0.5

A colorimeter contains 400 g of water at a temperature of 5^(@)C . Then, 200 g of water at a temperature of +10^(@)C and 400 g of ice at a temperature of -60^(@)C are added. What is the final temperature of the contents of calorimeter? Specific heat capacity of water -1000 cal //kg//K Specific latent heat of fusion of ice =80xx1000 cal //kg Relative specific heat of ice =0.5

If 2g ice at 0^(0)C is dropped into 10g water at 80^(0)C then temperature of mixture is

5 g of ice at 0^(@)C is dropped in a beaker containing 20 g of water at 40^(@)C . The final temperature will be

5g of water at 30^@C and 5 g of ice at -29^@C are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal//g-^@C and latent heat of ice =80 cal//g .

5g of water at 30^@C and 5 g of ice at -29^@C are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal//g-^@C and latent heat of ice =80 cal//g .

The water of mass 75 g at 100^(@)C is added to ice of mass 20 g at -15^(@)C . What is the resulting temperature. Latent heat of ice = 80 cal//g and specific heat of ice = 0.5 .