Arun
Last Activity: 5 Years ago
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
