if 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is

1 g of ice is mixed with 1 g of steam. At thermal equilibrium, the temperature of the mixture is¬

Assertion : If one gram of ice is mixed with 1 g of water at 80^(@)C , then the final temperature of mixture is zero. Reason : Latent heat of ice 80^(@) cal//g

100g ice at 0^(@)C is mixed with 100g water at 100^(@)C . The resultant temperature of the mixture is

100g ice at 0^(@)C is mixed with 100g water at 100^(@)C . The resultant temperature of the mixture is

1 gram of ice is mixed with 1 gram of steam. At thermal equilibrium, the temperature of the mixture is

1 gram of ice is mixed with 1 gram of steam. At thermal equilibrium, the temperature of the mixture is

10 g of ice at 0^@C is mixed with 100 g of water at 50^@C . What is the resultant temperature of mixture

100g of ice at 0C is mixed with 100g of water 80C The final temperature of the mixture will be