`10 gm` of ice at `-20^(@)C` is dropped into a calorimeter containing `10 gm` of water at `10^(@)C`, the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain:

50 g ice at 0^(@)C is dropped into a calorimeter containing 100 g water at 30^(@)C . If thermal . capacity of calorimeter is zero then amount of ice left in the mixture at equilibrium is

50 g ice at 0^(@)C is dropped into a calorimeter containing 100 g water at 30^(@)C . If thermal . capacity of calorimeter is zero then amount of ice left in the mixture at equilibrium is

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:

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:

How many gm of ice at -20^(@)C are needed to cool 200 gm of water from 25^(@)C to 10^(@)C ?

40 gm of water at 70^@C is poured into a calorimeter of mass 160 gm, which is at 20^@C . If the final temperature of the contents is 40^@C , what is the specific heat of the calorimeter? Specific heat of water = 4200 J /kg/C

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 -