In a insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Find the final temperature of the mixture (in kelvin.)
Given, `L_(fusion) = 80 cal//g = 336 J//g`
`L_("vaporization") = 540 cal//g = 2268 J//g`
`s_(ice) = 2100 J//kg.K = 0.5 cal//g.K`
and `s_("water") = 4200 J//kg.K = 1cal//g.K` .

Answer

273

Explanation

`0.05kg" steam at " 373K=overset(Q_(1))to0.05` kg water at 373 K
`0.05kg " water "373Koverset(Q_(2))to0.05` kg water at 237 K
0.45 kg ice at `253Koverset(Q_(3))to0.45` kg ice at 273 K
`0.45kg " ice at " 273Koverset(Q_(4))to0.45` kg water at 273 K
`Q_(1)=(50)(540)=27000 " cal"=27 "kcal" " " ,Q_(2)=(50)(1)(100)=5000 " cal"=5"kcal"`
`Q_(3)=(450)(0.5)(20)=4500" cal"=4.5 "kcal" " ",Q_(2)=(50)(100)=5000 "cal"=5" kcal"`
Now , since `Q_(1)+Q_(2)gtQ_(3)" but "Q_(1)+Q_(2)ltQ_(3)+Q_(4)`
ice will come to 273 K from 253 K, but whole ice will not melt. Therefore, temperature of the mixture is 273 K.