The minimum amount of 100-degree-c steam needed to melt 1 gram of 0-degree-celsius ice is

The mass of ice to be melted is

1 g = 10⁻³ kg.

Let m = the mass of steam required to melt the ice.

For water,

The latent heat of vaporization is 2260 kJ/kg

The latent heat of fusion is 334 kJ/kg

The steam should give up its latent heat of vaporization to melt the ice, which gives up its latent heat of fusion.

Therefore

(m kg) * (2260 kJ/kg) = (10⁻³ kg) * (334 kJ/kg)

m = 1.478 x 10⁻⁴ kg = 0.1478 g

Answer: 0.148 g