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Q82 (NDA-II/2009) Science & Technology › Basic Science (Physics, Chemistry, Biology) › Thermal physics Answer Verified

gm of ice at 0°C is mixed with 1 gm of steam at 0°C. After thermal equilibrium, the temperature the mixture is

Result
Your answer: —  ·  Correct: D
Explanation

To find the equilibrium temperature, we compare the heat required to melt ice and raise its temperature with the heat released by steam. To convert 1 g of ice at 0°C to water at 100°C, the heat required is the sum of the latent heat of fusion (80 cal/g) and the heat to raise the temperature of water by 100°C (100 cal), totaling 180 cal [t1, t3, t10]. Conversely, 1 g of steam at 100°C releases 540 cal just by condensing into water at 100°C [t3, t8]. Since the heat available from steam condensation (540 cal) is significantly greater than the heat needed to bring the ice to 100°C (180 cal), only a fraction of the steam will condense. The excess heat ensures the entire mixture reaches and remains at 100°C, with some steam remaining in the gaseous phase [c2, t8].

Sources

  1. [1] Physical Geography by PMF IAS, Manjunath Thamminidi, PMF IAS (1st ed.) > Chapter 22: Vertical Distribution of Temperature > Latent Heat > p. 294
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