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InterviewSolution
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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` . |
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Answer» Correct Answer - B::C Heat lost by steam during condensation `=M_(S)L_(V)` `Q_(1)=0.05xx2268xx1000=113400J` Heat lost by water of steam to cool to `273 K` `Q_(2)=0.05xx4200xx100=21,000 J ` `:.` Total heat available from steam `Q=134400J` Heat gained by ice to raise temp from `253k` to `273 k = Q_(1)^(1)=0.45xx2100xx(273-253)=18900 J` Heat gained by ice during metting at `273 k = Q_(2)^(1)` `Q_(1)^(2)=0.45xx336xx9000=151200 J` Total heat required `Q_(1)^(1) to Q_(1)^(2)=Q^(1)=170100 J` `Q^(1)gtQ` `:.` Whole of ice will not melt. Equilibrium temp of mixture `=273 k =0^(@)C` |
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