One gram of steam at hundred degree celsius melts how much ice at 0 degree celsius
Answers
Explanation:
The heat exchange must be calculated in several steps:
1. The heat lost by the steam to the ice when the steam condenses
2. The heat gained by the ice when it melts
3. The heat lost by the condensed steam to the original water as the temperature of the condensed steam equalizes with that of the water.
Part 1: The heat released by the steam as it condenses:
ΔH1 = (mass of steam)(latent heat of vaporization) = (0.010kg)(2260kJ/kg)
= 22.60 kJ
Part 2: Heat needed to melt ice:
ΔH2 = (mass of ice)(latent heat of fusion)=(0.050kg)(333 kJ/kg)
= 16.65 kJ
Since heat lost by steam must equal heat gained by water, an additional amount of (22.60-16.65)=5.95 kJ is absorbed by the melted ice and raises the temperature of the water, but is not enough to bring the water to 100° so it won’t boil. [Because (0.050kg)( (4.186 kJ/kg-K)(100°C) > 16.65 kJ]
Heat lost by the 100°C water = heat gained by the 0° water,
Or ΔHcold = -ΔHhot
(0.050kg)(333 kJ/kg)+ (0.050kg)(4.186 kJ/kg-°C)(Tf-0°C) =
[(0.010kg)(2260kJ/kg) + (0.010kg)(4.186 kJ/kg-°C)( 100°C - Tf)]
(16.65kJ + 0.2093 Tf – 0kJ) =( 22.60kJ + 4.186kJ - 0.04186 Tf)
0.25116 Tf = 10.136, Steam at 100 °C is added to ice at 0 °C. Find the amount of ice melted and the final temperature when the mass of steam is 10.0 g and the mass of ice is 50.0 g. Freezing temperature of water: 0...
Steam at 100 °C is added to ice at 0 °C. Find the amount of ice melted and the final temperature when the mass of steam is 10.0 g and the mass of ice is 50.0 g.
Freezing temperature of water: 0 degrees c
Boiling temperature of water: 100 degrees c
Latent heat of fusion of water: 333 kJ/kg
Latent heat of vaporization of water: 2260 kJ/kg
Specific heat of water: 4186 J/kgK
Specific heat of ice: 2220
EXPERT ANSWERS
T-NEZ | CERTIFIED EDUCATOR
The heat exchange must be calculated in several steps:
1. The heat lost by the steam to the ice when the steam condenses
2. The heat gained by the ice when it melts
3. The heat lost by the condensed steam to the original water as the temperature of the condensed steam equalizes with that of the water.
Part 1: The heat released by the steam as it condenses:
ΔH1 = (mass of steam)(latent heat of vaporization) = (0.010kg)(2260kJ/kg)
= 22.60 kJ
Part 2: Heat needed to melt ice:
ΔH2 = (mass of ice)(latent heat of fusion)=(0.050kg)(333 kJ/kg)
= 16.65 kJ
Since heat lost by steam must equal heat gained by water, an additional amount of (22.60-16.65)=5.95 kJ is absorbed by the melted ice and raises the temperature of the water, but is not enough to bring the water to 100° so it won’t boil. [Because (0.050kg)( (4.186 kJ/kg-K)(100°C) > 16.65 kJ]
Heat lost by the 100°C water = heat gained by the 0° water,
Or ΔHcold = -ΔHhot
(0.050kg)(333 kJ/kg)+ (0.050kg)(4.186 kJ/kg-°C)(Tf-0°C) =
[(0.010kg)(2260kJ/kg) + (0.010kg)(4.186 kJ/kg-°C)( 100°C - Tf)]
(16.65kJ + 0.2093 Tf – 0kJ) =( 22.60kJ + 4.186kJ - 0.04186 Tf)
0.25116 Tf = 10.136, Tf = 40.36°C (final temperature)
Total heat gained by ice is equal to the total heat lost by steam.
For ice to completely convert into water, heat required is m1 Lf =1×80=80cal
For steam to completely convert into water, heat released is m2 Lv =1×540=540 cal
Hence, first 80 calories will not be enough for the steam to condense completely.
Now, to convert melted water to 100° C from 0° C, heat required is m1s(100−0)=1×1×100=100cal
So, total energy required to heat ice to water 100° C is 100+80=180 cal.
Hence, even this amount of energy is not enough for the steam to condense completely. Hence, the final temperature of the mixture will be 100° C.
Note- finally the mixture will consist of both steam and water at 100° C.
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