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`1kg` ice at `0^(@)C` is mixed with `1kg` of steam at `100^(@)C` what will be the composition of the system when thermal equilibrium is reached ? Latent beat of fusion of ice `= 3.36xx 10^(5)J kg^(-1)` and latent head of vaporization of water `= 2.26 xx 10^(6)J kg^(-1)` |
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Answer» Correct Answer - A::C::D Heat obserbed by the ice to raise the temperature `100^(@)C` `Q_(1) = 1 xx 3.36 xx 10^(5) + 1 xx 4200 xx 100` ` = 3.36 xx 10^(5) + 4.2 xx 10^(5)` ` = (3.36 + 4.2) xx 10^(5)` ` = 7.56 xx 10^(5) = 0.756 xx 10^(6)` `Q_(2) ` heat released by steam ` = 1 xx 2.26 xx 10^(6)J` ` = 3.26 xx 10^(6)J` THe extra heat `= Q_(2) - Q_(1)` ` = (2.26 - 0.756 xx 10^(6)` `= 1.506 xx 10^(6)` THe arrount of steam condensed ` = (1.506 xx 10^(6))/(2.26 xx 10^(6))` `= 0.665 kg = 665 gm` The extra ice` = (1000 - 665) = 335gm` The amount of ice formed `= 1000 +335 = 1335g` |
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