Ice at -15 deg. Celsius and steam at 145 deg. Celsius are brought together at at

Question

Ice at -15 deg. Celsius and steam at 145 deg. Celsius are brought together at atmospheric pressure in a perfectly insulated container. After thermal equilibrium is reached, the liquid phase at 60 deg. Celsius is present. Ignoring the container and the equilibrium vapor pressure of the liquid at 60 deg. Celsius, Find the ratio of the mass of steam to the mass of ice. (Spec. heats are ice 2x10^3J/kgdegCelsius, water = 4186 J/kg deg. Celsius, Steam = 2020 J/kg deg Celsius, Latent heats of fusion =333 kJ/kg, of vaporization = 2260 kj/kg)

Explanation / Answer

mi = mass of ice

ms = mass of steam

Heat required to heat mi from -15 to 0

specific heat of ice = 2.06 kJ/kgC

E = 2.06 kJ/kgC x mi x 15 = 30.9 mi kJ

Heat required to melt ice
heat of fusion of ice = 334 kJ/kg

E = 334 kJ/kg x mi = 334 mi kJ

Heat required to heat mi from 0 to 60
specific heat of water = 4.2 kJ/kgC

E = 4.2 kJ/kgC x mi x 60 = 252 mi kJ

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Heat given off cooling steam from 145 to 120
specific heat of steam = 2.1 kJ/kgK
E = 2.1 kJ/kgK x ms x 25 = 52.5 ms kJ

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

heat given off by cooling water 120C to 60C
specific heat of water = 4.2 kJ/kgC
E = 4.2 kJ/kgC x ms x 60 = 252 ms kJ

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mi(31+334+252) = ms(52.5+2260+252)

617 mi = 2565 ms

ms/mi = 0.24

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