Question

If 250 g of water at 285 K is mixed with 80 g of ice at –150C. Calculate the final
temperature and composition of mixture.
[Given –
Heat capacity of ice - 0.5 cal/g-K
Heat capacity of water - 1 cal/g-K
Heat of fusion - 334 J/g
And latent heat of vaporization = 2260 J/g]

Answer 1

Answer:

Hope it helps you.

Answer 2

Given info : If 250 g of water at 285 K is mixed with 80 g of ice at –150°C.]

To find : the final temperature and the composition of mixture are ...

solution : 250g of water is at 285K or (285 - 273) = 12°C

if water loses its heat at 0°C ,

  heat , Q₁ = msΔT = 250g × 1 cal/g/°C × 12°C = 3000 cal

if ice increase its temperature upto 0°C,

heat needed, Q₂ = m's'ΔT' = 80g × 0.5 cal/g/°C × 150°C = 6000 cal

here it is clear that Q₁ < Q₂

it means, the final temperature of mixture would be 0°C.

energy lost by water to convert into ice, Q₃ = mLf  = 250g × 80 cal/g = 20000 cal

Q₁ + Q₃ > Q₂ , it means some part of water converted into ice and some remained in its form.

let x g of water is converted into ice.

   Q₁ + x g × 80 cal/g = Q₂

⇒ 3000 cal + 80x = 6000 cal

⇒ 80x = 3000

⇒ x = 37.5g

therefore the composition of mixture ;

ice = 80g + 37.5 g = 117.5 g

water = 250 - 37.5 = 212.5 g