Question

Ice at −14.0 °C and steam at 142 °C are brought together at atmospheric pressure in a perfectly insulated container. After thermal equilibrium is reached, the liquid phase at 50.0 °C is present. Ignoring the container and the equilibrium vapor pressure of the liquid, find the ratio of the mass of steam to the mass of ice. The specific heat capacity of steam is 2020 J/(kg.C°).

Answer 1

let mi be mass of ice, ms be mass of steam

heat required to heat mi from -14 to 0
specific heat of ice is 2.06 kJ/kgC
E = 2.06 kJ/kgC x mi x 14 = 28.84 mi kJ

heat required to melt ice
heat of fusion of ice is 334 kJ/kg
E = 334 kJ/kg x mi = 334 mi kJ

heat required to heat mi from 0 to 50
specific heat of water is 4.2 kJ/kgC
E = 4.2 kJ/kgC x mi x 50 = 210 mi kJ



heat given off cooling steam from 142 to 100
specific heat of steam is 2.1 kJ/kgK
E = 2.1 kJ/kgK x ms x 42= 88.2 ms kJ

heat given off by condensing steam
heat of vaporization of water is 2260 kJ/kg
E = 2260 kJ/kg x ms = 2260 ms kJ

heat given off by cooling water 100C to 50C
specific heat of water is 4.2 kJ/kgC
E = 4.2 kJ/kgC x ms x 50 = 210 mi kJ



Now we can set the two totals equal to each other.

mi(29+334+210) = ms(88+2260+210)
573mi = 2558ms

you want the ratio
ms/mi = 573/2558 = 0.224

Hope this helps. please thumbs up?