The volume of steam produced by `1g` of water at `100^(@)C` is `1650 cm^(3)`. Calculate the change in internal energy during the change of state. Given `J= 4.2xx10^(7) erg`.
`cal.^(-1), d= 981 cm^(s-2)`. Latent heat of stream `= 540 cal. G^(-1)`

The latent heat of vaporisation of water at 100 ^(@)C is 540 cal g^(-1) . Calculate the entropy increase when one mole of water at 100^(@)C is evaporated

10 gram of water at 45^(@)C is added to 5 gram of ice at -30^(@)C . Find the final temperature of the mixture. Specific heat of ice is 0.55 cal g^(-1)""^(@)C^(-1) and that of water is 1.0 cal g^(-1)""^(@)C^(-1) latent heat of fusion of ice is 80 cal g^(-1) ? (ii) To rise the temperature of 100 gram of water from 24^(@) to 90^(@)C by steam at 100^(@)C . Calculate the amount of steam required. Specific heat of water is 1.0 cal g^(-1)""^(@)C^(-1) . Latent heat of steam is 540 Cal kg^(-1) .

Determine the change in enthalpy and internal energy when 1 mole of water completely vaporises at 100^(@)C and 1 atm pressure. [Latent heat of vaporisation of water=536 cal*g^(-1) ]

1gm water at 100^(@)C is heated to convert into steam at 100^(@)C at 1atm . Find out chage in internal energy of water. It is given that volume of 1gm water at 100^(@)C = 1c c , volume of 1gm steam at 100^(@)C = 167 1 c c . Latent heat of vaporization = 540 cal//g . (Mechanical equivalent of heat J = 4.2J//cal)

1gm water at 100^(@)C is heated to convert into steam at 100^(@)C at 1atm . Find out chage in internal energy of water. It is given that volume of 1gm water at 100^(@)C = 1c c , volume of 1gm steam at 100^(@)C = 167 1 c c . Latent heat of vaporization = 540 cal//g . (Mechanical equivalent of heat J = 4.2J//cal)

Steam at 100^(@)C is passed into 20 g of water at 10^(@)C when water acquire a temperature of 80^(@)C , the mass of water present will be [Take specific heat of water = 1 cal g^(-1).^(@) C^(-1) and latent heat of steam = 540 cal g^(-1) ]

Steam at 100^(@)C is passed into 20 g of water at 10^(@)C . When water axquires a temperature of 80^(@)C , the mass of water present will be [Take specific heat of water 1"cal"g^(-1)c^(-1) latent heat of steam = 540 cal//g^(-1) ]

Steam at 100^(@)C is passed into 20 g of water at 10^(@)C when water acquire a temperature of 80^(@)C , the mass of water present will be [Take specific heat of water = 1 cal g^(-1).^(@) C^(-1) and latent heat of steam = 540 cal g^(-1) ]