1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when Thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 10^5 J/Kg and latent heat of vaporization of water= 2.26 × 10^6 J/Kg.

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Arun · Answer

Dear Abhishek

let the final temperature be T°C

=> heat absorbed by 1 kg ice to melt to water at 0°C = 3.36 x 10^5 J

=> also heat potentially released by 1 kg steam to be water at 100°C = 2.26 x 10^6 J which is much more

Q = mcΔT

c = specific heat of water = 4186 J / kg K

since there is a large amount of heat available from the steam, let us calculate the heat needed for the 1 kg water at 0°C to become water at 100°C

=> total heat gain = 3.36 x 10^5 J + 1x4186 (100) = 336000 + 418600 = 754600 J

=> total heat loss = total heat gained = 754600

however heat energy available from condensation of the steam = 2260000

therefore fraction of steam condensed = 754600 / 2260000 = 0.334 of 1 kg = 334 g

therefore all the ice will melt to water at 0°C then increase its temperature to 100°C while only 0.334 kg steam condenses at 100°C and ~ 667 g remains steam. T = 100 °C

hope this helps

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1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when Thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 10^5 J/Kg and latent heat of vaporization of water= 2.26 × 10^6 J/Kg.

1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when Thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 10^5 J/Kg and latent heat of vaporization of water= 2.26 × 10^6 J/Kg.

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