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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

The correct Answer is:
`273K`
`Sigma DeltaQ = 0`
Heat lost by steam to convert into `0^(@)C` water
`H_(L) = 0.05 xx 540 +0.05 xx 100 xx1`
`= 27 +5 = 32 kcal`
Heat required by ice to change into `0^(@)C` water
`H_(g) = 0.45 xx (1)/(2) xx 20 +0.45 xx 80 = 4.5 +36.00 = 40.5 kcal`
Thus, final temperature of mixture is `0^(@)C = 273 K`
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