Multiple choice

In an insulated vessel, $0.05\ kg$ steam at $373\ K$ and $0.45\ kg$ of ice at $253\ K$ are mixed. Then, find the final temperature of the mixture. Given, $L_{fusion} = 80 cal/g = 336 J/g$, $L_{vaporisation} = 540 cal/g = 2268 J/g$, $S_{ice} = 2100 J/kg K = 0.5 cal/gK$ and $S_{water} = 4200 J/kg K = 1 cal /gK$

  1. $273\ K$
  2. $373\ K$
  3. $100\ K$
  4. $0\ K$
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A Correct answer
Explanation

Heat lost by steam (cooling to 100C + condensing + cooling to 0C) vs heat gained by ice (warming to 0C + melting). 0.05kg steam releases 0.05*2268000 + 0.05*4200*100 = 113400 + 21000 = 134400 J. 0.45kg ice requires 0.45*2100*20 + 0.45*336000 = 18900 + 151200 = 170100 J. Since heat required to melt all ice is greater than heat released by steam, the final state is a mix of ice and water at 273 K.

AI explanation

Assess the heat exchange by calculating the total heat released if the steam completely condenses to water at 273 K and the total heat needed to bring the ice to 273 K and melt it. The steam releases 336,000 J (from 0.05 kg * 2268 J/g) upon condensation plus 21,000 J cooling to 273 K, totaling 357,000 J available. The ice requires 18,900 J to warm up and 37,800 J to melt, totaling only 56,700 J; since the available heat vastly exceeds the ice's needs, the final state is a mixture of water and melting ice in equilibrium. The final temperature of the mixture is 273 K.

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