1.

1 g steam at 100^(0)C is passed in an insulating vessel having 1 g ice at 0^(0)С. Find the equllibrium combosition of the mixture. Neglecting heat capacity of the vessel.

Answer»

Solution :Heat AVAILABLE from steam (changes from steam to water) `= ML = 1 xx 540 = 540` cal
Het gained by ice to change into water and then rise its temperature to `100^(0)C = m_("ice")L + m_("wa") cDeltaT = 1 xx 80 + 1 xx 1 xx (100 - 0) = 180` cal.
The above calculations show that some part of steam will CONDENSE to change the ice into water at `100^(0)C`. Let m be the mass of steam CONDENSED, then
`m xx 540 = 180 or m = (180)/(540)=1/3g` , Final CONTENTS : ice = 0g
water `= 1 + 1/3 = 4/3 g` steam `= 1 - 1/3=2/3g`.


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