Question

Calculate the amount of ice which is required to cool 150 g of water contained in a vessel of mass 100 g at 30C, such that the final temperature of the mixture is 5C. Take specific heat capacity of material of vessel as 0.4 J g1C-1 ; specific latent heat of fusion of ice = 336 J g-1; specific heat capacity of water = 4.2 J g-1C-1.

Answer 1

Answer:

Explanation:

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

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