How much energy is required to change exactly 1 g of ice at -20°C to steam at 120 °C? [Specific heat capacity of water = 4200J kg-1

oC

-1; Specific latent heat of fusion of ice = 334,000 Jkg

-1, specific heat capacity of steam is 2020 J/(kg °C), Specific latent heat of vaporization of water = 2,260,000 J/kg, specific heat capacity of ice = 2100 J/(kg K)]

**Answer**:

The processes involved:

Ice (-20°C) -----> Ice (0°C) -----> Water (0°C) -----> Water (100°C)

-----> Steam (100°C) -----> Water (120°C)

Energy required:

Ice (-20°C) to Ice (0°C), Q1 = mcθ = (0.001)(2100)(20) = 42J

Ice (0°C) to Water (0°C), Q2 = mL = (0.001)(334000) = 334J

Water (0°C) to Water (100°C), Q3 = mcθ = (0.001)(4200)(100) = 420J

Water (100°C) to Steam (100°C), Q4 = mL = (0.001)(2260000) = 2260J

Steam (100°C) to Steam (120°C), Q5 = mcθ = (0.001)(2020)(20) = 40.4J

Total energy required

= Q1 + Q2 + Q3 + Q4 + Q5

= 42 + 334 + 420 + 2260 + 40.4 = 3096.4J