`1g` of ice at `0^@ C` is mixed `1 g` of steam at `100^@ C`. The mass of water formed is

1 g of ice at 0^@C is mixed with 1 g of steam at 100^@C . After thermal equilibrium is achieved, the temperature of the mixture is

1 g of ice at 0^@C is mixed with 1 g of steam at 100^@C . After thermal equilibrium is achieved, the temperature of the mixture is

1 kg of ice at 0^@C is mixed with 1 kg of steam at 100^@C (i) The equilibrium temperature is 100^@C (ii) The equilibrium temperature is 0^@C (iii) The contents of mixture are : 665 g steam, 1335 g water (iv) The contents of mixture are 800 g steam, 1200 g water

60 g of ice at 0^@C is mixed with 60 g of steam at 100^@C . At thermal equilibrium, the mixture contains (Latent heat of steam and ice are 540 g^-1 and 80 cal g^-1 respectively, specific heat of water = 1cal g^-1^@C^-1 )

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

6 gm of steam at 100^@ C is mixed with 6 gm of ice at 0^@ C . Find the mass of steam left uncondensed. (L_(f)=80 cal//g,L_(v)= 540cal//g, S_(water) = 1cal//g-^(@) C) .