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 -

120 g of ice at 0^(@)C is mixed with 100 g of water at 80^(@)C . Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/ g-.^(@)C . The final temperature of the mixture is

120 g of ice at 0^(@)C is mixed with 100 g of water at 80^(@)C . Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/ g-.^(@)C . The final temperature of the mixture is

How many gram of ice at -14^@C are needed to cool 200 gm of water from 25^@C to 10^@C . Specific heat of ice = 0.5 cal/gm ^@C and latent heat of ice = 80 cal/gm.

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.

10gm ice at -10^(@)C is put in 30gm water at 30^(@)C . Final temperature of mixture (in^(@)C) is : Given (specific heat capacity of ice & water are 0.5 cal / gm^(@)C respectively and Latent heat of ice = 80 cal/gm).Write your answer in nearest integer.