`4 gm` of steam at `100^(@)C` is added to `20 gm` of water at `46^(@)C` in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation `= 540 cal//gm`. Specific heat of water `=1 cal//gm-^(@)C`.

Latent heat of vaporisation of water is 540 cal/gm.Explain the meaning of the statement.

In a container of negligible heat capacity, 200 gm ice at 0^(@)C and 100gm steam at 100^(@)C are added to 200 gm of water that has temperature 55^(@)C . Assume no heat is lost to the surroundings and the pressure in the container is constant 1.0 atm . (Latent heat of fusion of ice =80 cal//gm , Latent heat of vaporization of water = 540 cal//gm , Specific heat capacity of ice = 0.5 cal//gm-K , Specific heat capacity of water =1 cal//gm-K) What is the final temperature of the system ?

In a container of negligible heat capacity, 200 gm ice at 0^(@)C and 100gm steam at 100^(@)C are added to 200 gm of water that has temperature 55^(@)C . Assume no heat is lost to the surrpundings and the pressure in the container is constant 1.0 atm . (Latent heat of fusion of ice =80 cal//gm , Latent heat of vaporization of water = 540 cal//gm , Specific heat capacity of ice = 0.5 cal//gm-K , Specific heat capacity of water =1 cal//gm-K) Amount of the steam left in the system, is equal to

In a container of negligible heat capacity, 200 gm ice at 0^(@)C and 100gm steam at 100^(@)C are added to 200 gm of water that has temperature 55^(@)C . Assume no heat is lost to the surrpundings and the pressure in the container is constant 1.0 atm . (Latent heat of fusion of ice =80 cal//gm , Latent heat of vaporization of water = 540 cal//gm , Specific heat capacity of ice = 0.5 cal//gm-K , Specific heat capacity of water =1 cal//gm-K) Amount of the steam left in the system, is equal to

In a container of negligible heat capacity, 200 gm ice at 0^(@)C and 100gm steam at 100^(@)C are added to 200 gm of water that has temperature 55^(@)C . Assume no heat is lost to the surrpundings and the pressure in the container is constant 1.0 atm . (Latent heat of fusion of ice =80 cal//gm , Latent heat of vaporization of water = 540 cal//gm , Specific heat capacity of ice = 0.5 cal//gm-K , Specific heat capacity of water =1 cal//gm-K) What is the final temperature of the system ?

Calculate the K_b for water given latent heat of vaporisation of water 540 Cal//gm .

When m gram of steam at 100^(@) C is mixed with 200 gm of ice at 0^(@)C . it results in water at 40^(@) C . Find the value of m in gram . (given : Latent heat of fusion ( L_f ) = 80 cal/gm, Latent heat of vaporisation ( L_v ) = 540 cal/gm., specific heat of water ( C_w )= 1 cal/gm/C)

When m gram of steam at 100^(@) C is mixed with 200 gm of ice at 0^(@)C . it results in water at 40^(@) C . Find the value of m in gram . (given : Latent heat of fusion ( L_f ) = 80 cal/gm, Latent heat of vaporisation ( L_v ) = 540 cal/gm., specific heat of water ( C_w )= 1 cal/gm/C)

When m gram of steam at 100^@ C is mixed with 200 gm of ice at 0^@ C. it results in water at 40^@ C . Find the value of m in gram . (given : Latent heat of fusion ( L_f ) = 80 cal/gm, Latent heat of vaporisation ( L_v ) = 540 cal/gm., specific heat of water ( C_w )= 1 cal/gm/C)

Four cubes of ice at -10^(@)C each one gm is taken out from the refrigerator and are put in 150 gm of water at 20^(@)C . The temperature of water when thermal equilibrium is attained. Assuming that no heat is lost to the outside and water equivalent of contaner is 46 gm . (Specific heat capacity of water = 1 cal//gm-^(@)C , Specific heat capacity of ice =0.5 cal//gm-^(@)C , Latent heat of fusion of ice = 80 cal//gm-^(@)C )