A piece of ice of mass 40 g is dropped into 200 g of water at 50°C. Calculate the final temperature of water after all the ice has melted. Specific heat capacity of water = 4200 J/kgoC , Specific latent heat of fusion of ice = 336 x 103 J/kg.
Let t be the final temperature.
Then heat Liberated by water = m x c x Δt
= [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.