A piece of ice of mass 40g is added to 200g of water at 50 degree celsius.Calculate the final temperature of water when all the ice has melted.Specific heat capacity of water =4200 J per kg per K and specific latent heat of fusion of ice=336*1000 J per kg.
Let t be the final temperature.
Then heat Liberated by water = m x c x Δtwhere m= mass
= [200 x 10-3 x 4.2 x 103 x (50 – t)]
= (42,000 – 840t)
Heat absorbed by ice to change into water at 0oC = 40 x 10 x 336 x 103 = 13,440 J.
Heat absorbed by water to change its temperature from 0oC to toC = (40 x 10-3 x 4.2x 103 x t) = 168t
Total heat absorbed by water = (13,440 + 168t) J
According to the principle of calorimetry,
Heat given = Heat taken
or 42000 – 840t = 13440 + 168t
or 42000 – 13440 = 168t + 840t
or 1008t = 28560
or t = 28560/1008 = 28.33oC
Hence, the final temperature of water is 28.33oC