Question

a piece of ice of mass 40 gram is added to 200 gram of water at 50 degree Celsius. Calculate the final temperature of water when all the ice as milk tits. Specific heat capacity of water= 4200 J per kg per Kelvin and specific latent heat of fusion of ice = 336 × 10^3 Jkg^-1

Answer 1

totol heat gained by ice = total heat lost by water

Answer 2

⛦Hᴇʀᴇ Is Yoᴜʀ Aɴsᴡᴇʀ⚑
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Let t be the final temperature.

➧ Then heat Liberated by water is
➾ 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...✊