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Q 7)

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.


Solution



Let t be the final temperature.

Then heat Liberated by water = m x c x Δt           

where m= mass 
c = specific heat capacity
∆t change in temeperature

= [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


Physics

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