Question

A piece of mass 40g is added to 200g of water at 50°c.calculate the final temperature of water when all the ice has melted.specific heat capacity of


water=4200J/kg/K and specific latent heat of fusion of ice=336×10³J/kg.

Answer 1

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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

♦ 42000 – 840t
➾ 13440 + 168t

♦ 42000 – 13440
➾ 168t + 840t

♦ 1008t = 28560

t = 28560 / 1008
➾ 28.33oC ...✔

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Thanks...✊