If one kilogram water at `100^(@)C` is vapourised in open atmosphere. The correct statement is

Calculate the work done (w) and internal energy change ( DeltaU ), when one mole of water at 100^@C vaporises against an atmospheric pressure of 1 atm assuming ideal gas behaviour. Heat of vapourisation of water at 100^@C is 1020 cal mol^(-1) .

One kilogram of ice at 0^(@)C is mixed with one kilogram of water at 80^(@)C . The final temperature of the mixture is (take: specific heat of water=4200 J" "kg^(-1)K^(-1) , latent theat of ice =336" kJ "kg^(-1) )

One kilogram water at 0^(@)C is brought into contact with a heat reservoir at 100^(@)C . Find

One kilogram water at 0^(@)C is brought into contact with a heat reservoir at 100^(@)C . Find

One kilogram of ice at 0^(@)C is mixed with one kilogram of water at 80^(@)C . The final temperature of the mixture is (Take : specific heat of water =4200 J kg^(-1) K^(-1) , latent heat of ice = 336 kJ//kg^(-1) )

A 10.0 kg block of ice has a temperature of -10.0^(@)C . The pressure is one atmosphere. The block absorbs 4.11 xx 10^(6)J of heat. What is the final temperature of the liquid water ?

Consider the vaporization of 1g of water at 100^(@)C at one atmosphere pressure. Compute the work done by the water system in the vaporization and change internal energy of the system.

540 calories of heat convert 1 cubic centimeter of water at 100^oC into 1671 cubic centimeter of steam at 100^oC at a pressure of one atmosphere. Then the work done against the atmospheric pressure is nearly