Question

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 1

Correct Answer - `273K`
`0.05` kg steam at
`373 K overset(Q_(1))to 0.05 ` kg water at `373K`
`0.05` kg water at
`273 K overset(Q_(2))to 0.05` kg water at `273K`
`0.45` kg ice at
`273 K overset(Q_(3))to 0.45` kg ice at `273K`
`0.45` kg ice at
`273 K overset(Q_(4))to 0.45` kg at `273K`
`Q_(1) = (50)(540) = 27,000 "cal" = 27 "kcal"`
`Q_(2) = (5)(1)(100) = 5000"cal" = 5"kcal"`
`Q_(3)= (450)(0.5)(20) = 4500 "cal" = 4.5 "kcal"`
`Q_(4) = (450)(80) = 36000 "cal" = 36 "cal"`
Now since `Q_(1) + Q_(2) gt Q_(3) "but" Q_(1) + Q_(2) lt Q_(3) + Q_(4)` ice will come to `273K` from `253K` , but whole ice will not melt. Therefore , temperature of the mixture is `273K`.