A piece of ice of mass 0.04 kg is added to 0.2 kg of water at 50 °C. Calculate final temperature of water when all ice has melted.
Sp. Heat capacity of water =4200 J/kg°C
Sp. Latent heat of fusion of ice =336*10^3 J/kg.
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⛦Hᴇʀᴇ Is Yoᴜʀ Aɴsᴡᴇʀ⚑
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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 ...✔
__________
Thanks...✊
▬▬▬▬▬▬▬▬▬▬▬▬☟
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 ...✔
__________
Thanks...✊
mondalananya8765:
Hey.. Thanxx.. I was in dire need of this answer...
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