DAT General Chemistry: Heating and Cooling Curves

DAT High-Yield Concept

Slopes vs. Plateaus: Slanted regions represent single-phase heating where added thermal energy increases molecular kinetic energy ( $T$ rises). Horizontal plateaus represent phase co-existence where energy breaks intermolecular forces, changing potential energy ( $T$ remains constant).
Phase-Specific Heat Capacity: Each physical state (solid, liquid, gas) has its own specific heat capacity ( $c$ ). Solid ice ( $\pu{2.09 J/g degC}$ ), liquid water ( $\pu{4.184 J/g degC}$ ), and steam ( $\pu{2.01 J/g degC}$ ) require distinct values in $q = m c \Delta T$ .
Plateau Lengths: Enthalpy of vaporization ( $\Delta H_{\text{vap}}$ ) is substantially larger than enthalpy of fusion ( $\Delta H_{\text{fus}}$ ) because completely separating molecules into a gas requires overcoming all intermolecular attractions, making the vaporization plateau far wider.
Total Heat Calculations: Multi-step temperature and phase changes must be calculated separately for each segment and summed: $q_{\text{total}} = \sum q_{\text{single phase}} + \sum q_{\text{phase change}}$ .

Fundamentals of Heating and Cooling Curves

A heating curve graphically illustrates how the temperature of a substance changes under constant pressure (isobaric conditions) as heat energy is supplied at a steady rate. A cooling curve depicts the reverse process as heat is steadily removed.

Heating one gram of water from ice at -25 C to steam at 125 C — Heating one gram of water at a steady rate gives a curve of five stretches. On the sloping parts only one phase is present and the heat raises the temperature, by an amount set by the specific heat of that phase; liquid water has the largest specific heat, so its slope is the shallowest. On the two flat parts the temperature does not move at all, because the heat being supplied is going into separating the particles rather than speeding them up, and the sample holds at 0 degrees Celsius until all the ice has melted and at 100 degrees Celsius until all the water has boiled away. The boiling plateau is much the longer of the two, since vaporizing a gram of water takes 2260 J against the 334 J that melting it takes. Cooling the same sample traces the identical curve in reverse.

Grant-funded product on heating curve (Source: C3bc-taaccct · CC BY 3.0 · Wikimedia Commons)

Temperature Changes Within a Single Phase

In single-phase regions (solid, liquid, or gas), supplied thermal energy converts directly into kinetic energy ( $E_k$ ), increasing molecular motion and causing temperature to rise. The heat absorbed is calculated using:

q = m c \Delta T

where $q$ is heat ( $\pu{J}$ or $\pu{kJ}$ ), $m$ is mass ( $\pu{g}$ ), $c$ is the phase-specific heat capacity ( $\pu{J/g degC}$ ), and $\Delta T = T_{\text{final}} - T_{\text{initial}}$ .

Key Takeaway: Slope and Specific Heat

The slope of a single-phase heating line is inversely proportional to its specific heat capacity ( $\text{Slope} \propto 1/c$ ). A smaller specific heat capacity means a steeper temperature increase per unit of added heat.

Constant-Temperature Plateaus During Phase Transitions

During phase changes, adding heat does not increase temperature. Instead, thermal energy alters molecular potential energy by overcoming or reforming intermolecular forces (IMFs). Two phases coexist in dynamic equilibrium across these isothermal plateaus.

To quantify heat during phase transitions, enthalpy of transition is utilized:

q = m \Delta H_{\text{trans}} \quad \text{or} \quad q = n \Delta H_{\text{trans}}

Melting/Freezing Plateau: Uses enthalpy of fusion ( $\ce{H2O(s) -> H2O(l)}$ , $\Delta H_{\text{fus}} = \pu{334 J/g}$ or $\pu{6.01 kJ/mol}$ ).
Boiling/Condensation Plateau: Uses enthalpy of vaporization ( $\ce{H2O(l) -> H2O(g)}$ , $\Delta H_{\text{vap}} = \pu{2260 J/g}$ or $\pu{40.7 kJ/mol}$ ).

Worked Specific Heat Calculation

Problem: Calculate the total heat energy in kilojoules ( $\pu{kJ}$ ) required to convert $\pu{36.0 g}$ of ice at $\pu{-10.0 degC}$ into steam at $\pu{110.0 degC}$ at $\pu{1 atm}$ .

Constants for $\ce{H2O}$ : $c_{\text{ice}} = \pu{2.09 J/g degC}$ , $\Delta H_{\text{fus}} = \pu{334 J/g}$ , $c_{\text{liquid}} = \pu{4.184 J/g degC}$ , $\Delta H_{\text{vap}} = \pu{2260 J/g}$ , $c_{\text{steam}} = \pu{2.01 J/g degC}$ .

Step-by-Step Dimensional Analysis:

Warm ice ( $\pu{-10.0 degC}$ to $\pu{0.0 degC}$ ):
$q_1 = (\pu{36.0 g})(\pu{2.09 J/g degC})(\pu{10.0 degC}) = \pu{752.4 J} = \pu{0.752 kJ}$
Melt ice at $\pu{0.0 degC}$ :
$q_2 = (\pu{36.0 g})(\pu{334 J/g}) = \pu{12024 J} = \pu{12.02 kJ}$
Heat liquid water ( $\pu{0.0 degC}$ to $\pu{100.0 degC}$ ):
$q_3 = (\pu{36.0 g})(\pu{4.184 J/g degC})(\pu{100.0 degC}) = \pu{15062.4 J} = \pu{15.06 kJ}$
Vaporize water at $\pu{100.0 degC}$ :
$q_4 = (\pu{36.0 g})(\pu{2260 J/g}) = \pu{81360 J} = \pu{81.36 kJ}$
Heat steam ( $\pu{100.0 degC}$ to $\pu{110.0 degC}$ ):
$q_5 = (\pu{36.0 g})(\pu{2.01 J/g degC})(\pu{10.0 degC}) = \pu{723.6 J} = \pu{0.724 kJ}$
Sum all steps:
$q_{\text{total}} = \pu{0.752 kJ} + \pu{12.02 kJ} + \pu{15.06 kJ} + \pu{81.36 kJ} + \pu{0.724 kJ} = \pu{109.9 kJ}$

DAT Exam Trap Alert!

Never substitute $\Delta T = T_{\text{final}} - T_{\text{initial}}$ into a single overall equation. Phase transitions occur at constant temperature ( $\Delta T = 0$ ), making single $q = m c \Delta T$ calculations invalid across phase boundaries! Also, check unit compatibility before adding Joules to kilojoules.

Cooling Curves and Supercooling

Removing heat in an exothermic path decreases molecular kinetic energy until condensation or freezing occurs. During freezing, a liquid can temporarily drop below its freezing point without solidifying—a phenomenon called supercooling. Once nucleation initiates, latent heat released rapidly returns the system to its true freezing point plateau.

Single-Phase vs. Phase Transition Summary

Feature	Single-Phase Region (Diagonal Slopes)	Phase Transition (Horizontal Plateaus)
Temperature Change	Varies ( $\Delta T \neq 0$ )	Constant ( $\Delta T = 0$ , isothermal)
Energy Destination	Kinetic Energy ( $E_k$ changes)	Potential Energy ( $E_p$ breaks/forms IMFs)
Phases Present	Single phase (solid, liquid, or gas)	Two phases in dynamic equilibrium
Governing Equation	$q = m c \Delta T$	$q = m \Delta H_{\text{trans}}$ or $q = n \Delta H_{\text{trans}}$

Mnemonic:

References

Sources returned by Google Search grounding during lesson verification.

Heating Curve for Water - Chemistry LibreTexts — chem.libretexts.org
Heating and Cooling Curves - Chemistry LibreTexts — chem.libretexts.org

Heating and Cooling Curves

Summary

Practice Questions

AI Tutor