To calculate the grams of ice formed, we need to consider the phase changes involved: steam to water and water to ice. The specific heat capacity and latent heat of fusion for water must be taken into account.

1. Steam to Water:

- Initial temperature of steam: Approximately 100°C (boiling point of water)

- Final temperature of water: 100°C (boiling point of water)

- Latent heat of vaporization (steam to water): L_v = 2260 kJ/kg

The heat released when steam condenses to water at 100°C is given by:

Q_1 = m * L_v

where m is the mass of steam and Q_1 is the heat released.

2. Water to Ice:

- Initial temperature of water: 100°C (boiling point of water)

- Final temperature of ice: 0°C (freezing point of water)

- Specific heat capacity of water: c_w = 4.18 kJ/kg °C

- Latent heat of fusion (water to ice): L_f = 334 kJ/kg

The heat released when water cools from 100°C to 0°C and then freezes into ice is given by:

Q_2 = m * c_w * (100 - 0) + m * L_f

Since the heat released during the phase changes (steam to water and water to ice) is used to convert 100 grams of steam to ice, we can equate Q_1 and Q_2:

Q_1 = Q_2

m * L_v = m * c_w * (100 - 0) + m * L_f

Solving for the mass of steam (m):

m = (100 * c_w * 100 + 100 * L_f) / (L_v)

m = (41,800 kJ + 33,400 kJ) / (2260 kJ/kg)

m ≈ 33.85 kg

Therefore, 100 grams of steam will produce approximately 33.85 kilograms of ice when it turns into water and then ice.