0.05 kg steam at 373K and 0.45kg of ice at 253K are mixed in an insultated vessel. Find the equilibrium temperature of the mixture. Given, `L_(fusion)=80cal//g=336J//g, L_(vaporization)=540 cal//g=2268J//g,C_(ice)=2100J//KgK=0.5calg//gK and S_(water)=4200J//KgK=1cal//gK`

Answer

B, C

Explanation

(1) Heat lost by steam at `100^@C` to change to `100^@C` water
`mL_(vap)=0.05xx2268xx1000=1,13400J`
(2) Heat lost by `100^@C` water to change to `0^@C` water
`=0.05xx4200xx100=21,000J`
(3) Heat required by o.45kg of ice to change its temperature
from 253K to 273K
`=mxxS_(ice)xxDeltaT=0.45xx2100xx20=18,900J
(4) Heat required by `0.45kg` of ice at 273K to convert into
`0.45kg` water at 273K
`=mL_(fusion)=0.45xx336xx1000=151,200J`
From the above data it is clear that the amount of heat
required by 0.45kg of ice at 253K to convert into 0.45kg of
water at `273K(1,70,100J)` cannot be provided by heat lost
by 0.05 kg of steam at 373K to convert into water at 273K.
Therefore the final temperature will be 273K or `0^@C.`