Mixing of Steam and Ice – Rankers Physics
Topic: Thermal Physics
Subtopic: Calorimetry

Mixing of Steam and Ice

1 g of steam at 100°C is mixed with 1 g of ice at 0°C, then resultant temperature of the mixture is
100°C
200°C
50°C
75°C

Solution:

Heat required to raise 1 g of ice at 0°C to water at 100°C is \(1 \times 80 + 1 \times 1 \times 100 = 180\text{ cal}\). Since 1 g steam releases \(540\text{ cal}\) upon complete condensation, only a fraction of the steam condenses, maintaining the final temperature at 100°C.

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