`1 kg` ice at `-10^(@)C` is mixed with `0.1 kg` of steam at `200^(@)C`. If final temperature of mixture at equilibrium is `T_(eq)=(58x)/11`, then fill the vallue of `x`
Latent heat of fusion of ice `=80` cal/gram, latent heat of vaporization of water `=540` cal/gram, specific heat capacity of ice `~=` specific heat of water `=0.5` cal/gram-K`

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) 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

1 kg of ice at 0^(@)C is mixed with 1.5 kg of water at 45^(@)C [latent heat of fusion = 80 cal//gl. Then

1 kg of ice at 0^(@)C is mixed with 1.5 kg of water at 45^(@)C [latent heat of fusion = 80 cal//gl. Then

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)