Question

A piece of ice of mass 40 gram is added to 200 gram of water at 50℃. calculate the final temperature when all the ice is melted???

Answer 1

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.