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

500g of ice at 0°C is mixed with 1g steam at 100°C . The final temperature of the mixture is