Calculate the amount of ice which is required to cool 150 g of water contained in a vessel of mass 100 g at 30°C, such that the final temperature of the mixture is 5oC. Take specific heat capacity of material of vessel as 0.4 J g-1 oC-1 ; specific latent heat of fusion of ice = 336 J g-1 ; specific heat capacity of water = 4.2 J g-1 °C-1.
Calculate the amount of ice which is required to cool 150 g of water contained in a vessel of mass 100 g at 30°C, such that the final temperature of the mixture is 5oC. Take specific heat capacity of material of vessel as 0.4 J g-1 oC-1 ; specific latent heat of fusion of ice = 336 J g-1 ; specific heat capacity of water = 4.2 J g-1 °C-1.
The correct option is C 46.92 g
Heat lost by the vessel and water
= [150 x 4.2 x (30 - 5) J + 100 x 0.4 x (30-5)] J
= (150 x 4.2 x 25 + 100 x 0.4 x 25) J
= (15,750 + 1000) J= 16,750 J
Heat gained by ice to change into water at 0 °C = m x specific latent heat of fusion of ice = {m x 336}
Heat gained by water to come to final temperature (5 °C)
= m x specific heat capacity of water x final temperature = (m x 4.2 x 5) = 21m
If there is no loss of heat to the surroundings, According to the principle of calorimetry:
Heat gained by ice to melt and come to final temperature = Heat lost by vessel and water
or 336m + 21m= 16,750
or 357m = 16,750
or m = 16750/357 = 46.92
46.92 g
Heat lost by the vessel and water
= [150 x 4.2 x (30 - 5) J + 100 x 0.4 x (30-5)] J
= (150 x 4.2 x 25 + 100 x 0.4 x 25) J
= (15,750 + 1000) J= 16,750 J
Heat gained by ice to change into water at 0 °C = m x specific latent heat of fusion of ice = {m x 336}
Heat gained by water to come to final temperature (5 °C)
= m x specific heat capacity of water x final temperature = (m x 4.2 x 5) = 21m
If there is no loss of heat to the surroundings, According to the principle of calorimetry:
Heat gained by ice to melt and come to final temperature = Heat lost by vessel and water
or 336m + 21m= 16,750
or 357m = 16,750
or m = 16750/357 = 46.92
