Question

) A piece of ice of mass 60 g is dropped into 140 g of water at 30°C.
Calculate the final temperature of water when all the ice has melted.
(Assume no heat is lost to the surrounding)
Specific heat capacity of water = 4.2 Jg kl 1
Specific latent heat of fusion of ice = 336 Jg​

Answer 1

Answer:

3.75 gms of ice and 196.25 gms of water at $0^{0}$C

Explanation:

Heat lost by water is = heat gained by ice

∴  Δm(80)=150(1)(30)

=>Δm=150($\frac{3}{8}$) = 56.25

∴ (60-56.25)gms of ice = 3.75

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Answer 2

HELLO DEAR,

let the final temperature of water is T.

now,

energy lost by water = (mass of water) × (specific heat of water) × (initial temperature - final temperature)

=> Mw × Cp × (30° - T)

=> 140 × 4.2 × (30° - T)----------( 1 )

also,

energy gain by ice = (mass of ice) × (latent heat of fusion + specific heat of water × final temperature)

=> M × (L + Cp×T)

=> 60 × (336 + 4.2 × T)----------( 2 )

by conservation of energy,

Mw × Cp × (30° - T) = M × (L + Cp×T)

=> 140 × 4.2 × (30 - T) = 60 × (336 + 4.2 × T)

=> 140 × 0.7 × (30 - T) = 10 × (336 + 4.2T)

=> 9.8 × (30 - T) = (336 + 4.2T)

=> 294 - 9.8T = 336 + 4.2T

=> 294 - 336 = 4.2T + 9.8T

=> -42 = 14T

=> T = - 3°

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THANKS