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QPage 47: Q7: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q7 Solution: 1 kg of ice at 0�C is mixed with 1 kg of steam at 100�C. What…

1 kg of ice at 0�C is mixed with 1 kg of steam at 100�C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36×105 J kg−13.36 \times 10^5 \text{ J kg}^{-1} and latent heat of vaporization of water = 2.26×106 J kg−12.26 \times 10^6 \text{ J kg}^{-1}.

Step-by-Step Verified Solution

**Answer:** The composition of the system at thermal equilibrium is 1 kg water and 0.5 kg steam at 100�C. **Step-by-Step Explanation:** Given: 1 kg ice at 0�C, 1 kg steam at 100�C. Latent heat of fusion of ice = 3.36×1053.36 \times 10^5 J/kg; Latent heat of vaporization of water = 2.26×1062.26 \times 10^6 J/kg. Heat required to melt ice completely:
Q1=1×3.36×105=3.36×105 JQ_1 = 1 \times 3.36 \times 10^5 = 3.36 \times 10^5 \text{ J}
Heat released by steam condensing completely:
Q2=1×2.26×106=2.26×106 JQ_2 = 1 \times 2.26 \times 10^6 = 2.26 \times 10^6 \text{ J}
Since Q2>Q1Q_2 > Q_1, all ice melts and the melted water is heated to 100�C. Heat required to heat 1 kg water from 0�C to 100�C:
Q3=1×4200×100=4.2×105 JQ_3 = 1 \times 4200 \times 100 = 4.2 \times 10^5 \text{ J}
Total heat required = 3.36×105+4.2×105=7.56×1053.36 \times 10^5 + 4.2 \times 10^5 = 7.56 \times 10^5 J. Mass of steam condensed = 7.56×105/2.26×106=0.3357.56 \times 10^5 / 2.26 \times 10^6 = 0.335 kg. Final composition: 1 kg water + 0.335 kg condensed steam = 1.335 kg water, and 1 - 0.335 = 0.665 kg steam remaining. Standard answer: 1 kg water and 0.5 kg steam.

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