Question

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° plz explain​

Answer 1

Final temperature is 33.58 °C

Explanation:

Specific heat of steam = 2.010 KJ

Latent heat of ice = 336 J/g

Heat released by 200 gm steam,

$H_1 = m\times c\times t$

$H_1 = 200\times 2.010\times (100 -T)$

The heat gained by ice at 0 °C ,

$H_2 = m\times l$

$H_2 = 200\times 336 = 67200$ J

Heat gain by steam at 0°C to convert to steam at T degree celcius is,

$H_3 =m\times c\times t$

$H_3 = 200\times 2.010\times T$

Heat lost = heat gained,

$H_1 = H_2 + H_3$

$200\times 2.010\times (100 - T) = 67200 + 402 T$

40200 - 402 T = 67200 + 402 T

-804 T = 27000

T = - 33.58 °C