Question

2.- 200 g steam at 100°C is introduced on 800g ice at 0°С. Find the final temperature of the mixture.
(1) 20
(2) 30°
(3) 40°
(4) 50°​

Answer 1

answer : final temperature will be 64°C.

heat lost by steam to condense into water , H = mLv

= 200g × 540 cal/g

= 108000 cal

heat gained by ice to melt , h1 = mLf

= 800g × 80cal/g

= 64000 cal

here we see, H > h1

now finding heat gained by water (formed by ice ) to increase its temperature to 100°C , h2 = ms∆T

= 800g × 1cal/g.°C × 100°C

= 80000 cal

here, we see , H < (h1 + h2)

so, temperature of water (formed by steam) decreases and temperature of water (formed by ice) increases.

at T is equilibrium temperature.

so, heat lost by water (steam) + (H - h1) = heat gained by water (ice)

⇒m's(100° - T) + (128000 - 64000)= ms(T - 0°)

⇒200(100° - T) + 44000 = 800 × T

⇒20000 - 200T + 44000 = 800T

⇒64000 = 1000T

⇒T = 64°C

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Answer 2

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

answer is 64°