Question

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

Answer 1

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