Heating Curves and Phase Changes explained the flat steps where the temperature stops rising while ice melts or water boils. Specific Heat gave the heat for the sloped parts. Put them together to find the heat for a whole trip, say from ice in the freezer to steam.
Key idea
Add one term per segment. Warming or cooling a phase: $q = mc\Delta T$, with that phase's own specific heat. A phase change at constant temperature: $q = n\Delta H$ (melting $\Delta H_\text{fus} = 6.01\ \text{kJ/mol}$, vaporizing $\Delta H_\text{vap} = 40.7\ \text{kJ/mol}$ for water). Water: $c_\text{ice} = 2.09$, $c_\text{liquid} = 4.184$, $c_\text{steam} = 2.01\ \text{J/(g·°C)}$.
Why it works
On a sloped segment, added energy speeds up the particles, raising the temperature. On a flat segment, it pulls particles apart against their attractions (melting loosens them; boiling separates them completely) without speeding them up, so the temperature stays put until the change is complete. Each phase stores energy differently, so each has its own specific heat; liquid water's is about twice ice's or steam's. Boiling takes almost seven times as much energy as melting, because the molecules must be separated completely.
Worked example: Ice to warm water
How much heat turns $50.0\ \text{g}$ of ice at $-10.0^\circ\text{C}$ into water at $25.0^\circ\text{C}$?
- Warm the ice to $0^\circ\text{C}$: $q_1 = (50.0)(2.09)(10.0) = 1045\ \text{J}$.
- Melt it: $n = 50.0/18.02 = 2.775\ \text{mol}$; $q_2 = 2.775 \times 6010 = 16{,}680\ \text{J}$.
- Warm the water to $25.0^\circ\text{C}$: $q_3 = (50.0)(4.184)(25.0) = 5230\ \text{J}$.
- Total: $1045 + 16{,}680 + 5230 = 22{,}950\ \text{J} = 23.0\ \text{kJ}$.
Answer: $23.0\ \text{kJ}$
Worked example: Hot water to steam
How much heat turns $25.0\ \text{g}$ of water at $80.0^\circ\text{C}$ into steam at $110.0^\circ\text{C}$?
- Warm the liquid to $100^\circ\text{C}$: $q_1 = (25.0)(4.184)(20.0) = 2092\ \text{J}$.
- Boil: $n = 25.0/18.02 = 1.387\ \text{mol}$; $q_2 = 1.387 \times 40{,}700 = 56{,}470\ \text{J}$.
- Warm the steam: $q_3 = (25.0)(2.01)(10.0) = 503\ \text{J}$.
- Total: $59{,}060\ \text{J} = 59.1\ \text{kJ}$.
Answer: $59.1\ \text{kJ}$
A cooling curve is the same path run backward, with every $q$ negative. Condensing 1 g of steam at $100^\circ\text{C}$ releases $40.7/18.02 = 2.26\ \text{kJ}$, more than five times the $0.418\ \text{kJ}$ released by cooling 1 g of liquid water from $100^\circ\text{C}$ to $0^\circ\text{C}$.
Common mistake
Using liquid water's specific heat (4.184) for ice and steam too, or leaving out the phase-change steps. The melting step alone was 73% of the first example's total. Check: list every segment the path crosses, one sloped or flat piece at a time, and make sure each phase uses its own $c$ and each change of state its own $\Delta H$, in joules.