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


  1. 28.33oC

  2. 38.88oC

  3. 25.33oC

  4. 45.55oC


Solution

The correct option is A

28.33oC


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.

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