Question

30 g of ice at 0 c and 20 g of steam at 100 c are mixed.The composition of the resultant mixture is

Answer 1

mass of water = 30g and mass of steam = 20g

heat lost by steam to condense, H1 = ml_v

= 20g × 540 cal/g

= 10800 cal

heat gained by ice to melt, H2 = m'l_f

= 30g × 80cal/g

= 2400 cal

and heat gained by water (ice) to increase the temperature into 100°C , H3 = 30g × 1 cal/g.°c × 100°C = 3000 cal

total heat gained by ice = H2 + H3 = 2400 + 3000 = 5400 cal < H1 = 10800 cal

definitely, steam wouldn't completely condense into water.

let m mass of steam is condense.

then, 5400 cal = m × 540

m = 10g

hence, mass of water in mixture = 10g + 20g = 30g

and mass of steam = 20g

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Answer 2

$\huge\bold\purple{Answer:-}$

heat lost by steam to condense, H1 = ml_v

= 20g × 540 cal/g

= 10800 cal

heat gained by ice to melt, H2 = m'l_f

= 30g × 80cal/g

= 2400 cal

and heat gained by water (ice) to increase the temperature into 100°C , H3 = 30g × 1 cal/g.°c × 100°C = 3000 cal

total heat gained by ice = H2 + H3 = 2400 + 3000 = 5400 cal < H1 = 10800 cal

definitely, steam wouldn't completely condense into water.

let m mass of steam is condense.

then, 5400 cal = m × 540

m = 10g

hence, mass of water in mixture = 10g + 20g = 30g

and mass of steam = 20g