Multiple choice

$1\;kg$ ice at $-20^{\circ}C$ is mixed with $1\;kg$ steam at $200^{\circ}C$. Then find equilibrium temperature and mixture content.

  1. $T=100^{\circ}C$, $\dfrac{20}{27}$ kg steam and $\dfrac{34}{27}$ kg water
  2. $T=0^{\circ}C$, $\dfrac{20}{27}$ kg steam and $\dfrac{34}{27}$ kg water
  3. $T=100^{\circ}C$, $\dfrac{34}{27}$ kg steam and $\dfrac{20}{27}$ kg water
  4. $T=0^{\circ}C$, $\dfrac{10}{27}$ kg steam and $\dfrac{38}{27}$ kg water
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A Correct answer
AI explanation

By the principle of calorimetry, heat lost by steam cooling to 100C and condensing at 100C equals heat gained by ice warming to 0C, melting at 0C, and the resulting water warming to 100C. Equating these heats yields 620 kcal lost by the steam and 100 kcal gained by the ice, causing the entire mass of 2 kg to become a mixture of water and steam at 100C. Specifically, the net heat available vaporizes water, leaving a final mixture of 34/27 kg of steam and 20/27 kg of water at 100 degrees Celsius.

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