Steam (water vapor) at 100 degrees C enters an insulated vessel containing 100 g of ice at 0 degrees C. How much steam must be introduced into the vessel so that equilibrium 50 g of the ice has melted and 50 g of the ice remains?

Specific heat of water = 4.184 J/K g

Latent heat of fusion of ice = 333 J/g

Latent heat of vaporization of water = 2260 J/g

1. I'll convert all mass measures into kg, energy into kJ.

2. Melting 0.05 kg of ice needs \(\displaystyle 0.05\ kg \cdot 333 \tfrac{kJ}{kg}\)

3. When steam condensates to water you have water of a temperature of 100°C. This hot water cools down to 0°C melting the ice too!

4. Solve for m (= mass of steam)

\(\displaystyle \underbrace{0.05\ kg \cdot 333 \tfrac{kJ}{kg}}_{\text{melting ice}} = \underbrace{m \cdot 2260 \tfrac{kJ}{kg}}_{\text{steam to water}} + \underbrace{m \cdot 4.184 \tfrac{kJ}{kg \cdot K} \cdot (100^\circ-0^\circ)K}_{\text{cool down water}}\)

5. I've got 0.0435 kg = 43.5 g