Q=M×Lfusion+M×S×(T2-T1)+M×Lvapourisation .Q=1×80cal+1×1×(100-0)+1×540cal.Q=80+100+540.Q=720cal.

Heat Required to Convert Ice to Steam:

Phase Changes:
- To convert ice at 0°C to water at 0°C, we need to provide heat for the phase change from solid to liquid. This is known as the heat of fusion.
- The heat of fusion for water is 334 J/g.

Heating Water:
- Once all the ice has melted into water at 0°C, we need to heat the water from 0°C to 100°C. This requires providing heat for an increase in temperature.
- The specific heat capacity of water is 4.186 J/g°C.
- The temperature difference is 100°C.

Phase Change to Steam:
- After heating the water to 100°C, we need to provide heat for the phase change from liquid to gas. This is known as the heat of vaporization.
- The heat of vaporization for water is 2260 J/g.

Calculations:
- Heat for melting ice: 1g x 334 J/g = 334 J
- Heat for heating water to 100°C: 1g x 4.186 J/g°C x 100°C = 418.6 J
- Heat for vaporizing water: 1g x 2260 J/g = 2260 J

Total Heat Required:
- Total heat = 334 J + 418.6 J + 2260 J = 3012.6 J
So, to convert one gram of ice at 0°C into steam at 100°C, we would need 3012.6 Joules of heat.