Question

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/kg^oC , Specific latent heat of fusion of ice = 336 x 10³ J/kg.

• A. 28.33^oC
• B. 38.88^oC
• C. 45.55^oC
• D. 25.33^oC

Answer 1

The correct option is A

28.33^oC

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 0^oC = 40 x 10 x 336 x 10³ = 13,440 J.

Heat absorbed by water to change its temperature from 0^oC to t^oC = (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 = 28.33^oC

Hence, the final temperature of water is 28.33^oC.