Question

How much energy is required to change a 45.0-g ice cube from ice at -6.0°C to steam at 115°C?

Answer 1

Specific heat capacities of water = 4.186 J/g/^oC

ice=2.093 J/g/^oC

steam=2.009 J/g/^oC

Latent heat of fusion for water = 333J/g

Latent heat of vapourization for water = 2260 J/g

Now energy required for conversion from -6 to 0 = mC = 45*2.093(6) = 565.11 J

For ice to water at 0 = mCf = 45*333=14985 J

water from 0 to 100 = 45*4.186*100 = 18837 J

for water to steam at 100 = mCv = 45*2260 = 101700 J

for steam from 100 to 115 = 45*2.009*6 = 542.43

Enet = 136629.54 J