Heat of vaporitzation physics

Question

Find what heat in calories is required to change 42 gg of 0∘C∘C ice to steam at 100∘C∘C. The heat of fusion Lf for water is 80 cal/gcal/g. The specific heat capacity of water is 1 cal/g⋅∘Ccal/g⋅∘C. The heat of vaporization for water Lv is 540 cal/gcal/g.

Answer 1

The solution to this problem involves 3 steps.

1)     Determine the heat necessary to convert 42 g ice at 0°C to water at 0°C using

ΔH   = Lf x Mass ice.

2)    Determine the heat necessary to raise 42g water to 100°C water using

ΔH = spec. heat H2O x mass water x (100 – 0)°C.

3)    Determine the heat required to vaporize 42 g water at 100 °C to steam at 100 °C using

ΔH = Lv x mass H2O.

Sum all the ΔHs to get the answer.