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In a insulated vessel, 0.05 kg steam at ...

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` .

Text Solution

Verified by Experts

`0.05Kg "steam at" 373K overset(Q_(1))rarr0.05kg "water at" 373K`
`0.05Kg "water at" 373K overset(Q_(2))rarr0.05kg "water at" 273K`
`0.05Kg "ice at" 373K overset(Q_(3))rarr0.45kg "ice at" 273K`
`0.05Kg "ice at" 373K overset(Q_(4))rarr0.45kg "water at" 273K`
`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_(4)=(450)(80)=36000 "cal"=36"kcal"`
Now sine `Q_(1)+Q_(2)gtQ_(3)` but `Q_(1)+Q_(2)ltQ_(4)` ice will come to `273K` from `253 K`, but whole ice will not melt, Therefore, temperature of the mixture is `273K`.
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