Phase Changes and Heating Curves

Phases of Matter

  • Low energy corresponds to the solid phase.
  • Medium energy corresponds to the liquid phase.
  • High energy corresponds to the gas phase.
  • Intermolecular forces (IMFs) are strong in solids, medium in liquids, and low in gases.

Heating Curve

  • A heating curve is a graph of temperature vs. heat added.
  • It can be generated by heating a substance (e.g., ice in a beaker) while measuring its temperature.
  • Initially, as heat is added, the temperature rises.
  • At the phase change (e.g., solid to liquid), the temperature remains constant.
  • The added heat energy is used to overcome the intermolecular forces (IMFs) holding the molecules together in the solid phase.
  • Once all of the solid has transitioned to liquid, the temperature will rise again as more heat is added.
  • The temperature remains constant again during the liquid to gas phase change.

Factors Affecting Heat Energy

  • When heating different substances to the same temperature (e.g., 90 degrees), the amount of energy required varies.
  • Mass matters: A larger mass requires more energy.
  • Substance matters: Solids may require more energy due to tightly packed particles.
  • Initial temperature matters: A substance starting at a lower temperature will require more energy to reach the final temperature.

Q equals MCAT

  • q=mcΔTq = mc \Delta T
  • This equation relates heat energy (q) to mass (m), specific heat capacity (c), and change in temperature (ΔT\Delta T).
    • q = heat energy (in joules)
    • m = mass (in grams)
    • c = specific heat capacity
    • ΔT\Delta T = change in temperature (final temperature - initial temperature)
  • Units are important for ensuring correct calculations.
  • Celsius can be used for temperature in the q equals mcat equation.

Example Problem

  • If the problem has specific heat, joules, or temperatures given, consider using q=mcΔTq = mc \Delta T.
  • Always list the given parameters, write out the equation, and plug in the values with units.
  • Make sure the units align/cancel out appropriately to arrive at the correct units for the answer (joules).

Numerical Example

  • Given: 34.4 grams of substance, initial temperature of 25°C, final temperature of 78.8°C, specific heat capacity of 2.44 J/g°C.
  • q=(34.4 g)×(2.44Jg⋅°C)×(78.8°C−25°C)q = (34.4 \text{ g}) \times (2.44 \frac{\text{J}}{\text{g} \cdot \text{°C}}) \times (78.8 \text{°C} - 25 \text{°C})
  • q=(34.4 g)×(2.44Jg⋅°C)×(53.8°C)q = (34.4 \text{ g}) \times (2.44 \frac{\text{J}}{\text{g} \cdot \text{°C}}) \times (53.8 \text{°C})
  • q=4515.03 Jq = 4515.03 \text{ J}
  • The answer should be expressed using the correct number of significant figures.