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

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