At 1 atm, how much energy is required to heat 43.0 g of H2O(s) at –22.0 °C to H2O(g) at 169.0 °C?

Answer 1

Do this in steps.
Within a single phase use
q = mass x specific heat x (Tfinal-Tinitial).
For example. For ice at -22.0C to ice at 0C, it will be
q = 43.0g H2O x specific heat ice x [0-(-22] = ?

At a phase change use
q = mass x heat fusion (at melting point) or
q = mass x heat vaporization (at boiling point).
For example.
At zero C, the melting point of ice,
q = 43.0g x heat fusion = ?

Another within the phase from zero C to 100 C for liquid water.
Another phase change at 100 from liquid to vapor.
Another within the phase (gas at 100) from 100 C to 169.0C (gas at 169.0).

Then add all of the q values together.