chem thermo stuff
Heating and Cooling Curves
Understanding that temperature does not always rise straight up with heat.
Common Misconception: The assumption that temperature increases linearly as heat is applied is incorrect.
Important Note: Pay attention to phase changes, where temperature remains constant despite heat addition.
Heating and Cooling Process
Initial State
Ice in the beaker starts below freezing (0 degrees Celsius).
Particles in a solid are organized but move slightly (depicted with vibration marks).
Heating of Ice (Slanted Line)
Graph shows a slanted line as ice heats from below 0 to 0 degrees Celsius.
Key Equation: Q = mcΔT, where:
Q = heat energy (Joules)
m = mass (grams)
c = specific heat capacity
ΔT = change in temperature (°C)
No melting occurs until ice reaches 0 degrees Celsius.
Melting Process (Flat Line)
At 0 degrees, ice begins to melt; line on the graph is flat (temperature does not increase).
Energy is used to break intermolecular bonds (latent heat of fusion).
Key Equation: Q = ΔH_sub_f * m, where ΔH_sub_f = heat of fusion (334 J/g).
The particles become less organized, moving freely without increasing temperature.
Heating of Water (Slanted Line)
After melting, the water's temperature increases above 0 degrees Celsius, with a new slanted line.
Still using Q = mcΔT, but with different c for liquid water (4.184 J/g°C).
Temperature rise continues until reaching boiling point (100 degrees Celsius).
Boiling Process (Flat Line)
Temperature holds at 100 degrees as water boils; energy is used to convert liquid to gas.
Key Equation: Q = ΔH_sub_v * m, where ΔH_sub_v = heat of vaporization (2260 J/g).
Particles move freely and energetically in vapor phase.
Final Phase (Heating Steam)
After boiling, if further heated, the steam's temperature will increase, leading to a slanted line again.
The complete heating/cooling cycle includes:
Solid (below 0) → melting (flat at 0) → liquid (0 to 100) → boiling (flat at 100) → gas (above 100).
Concepts of Energy Changes
Types of Energy:
Thermal Energy: Associated with temperature changes.
Phase Energy: Associated with state changes.
Energy transformations occur during temperature changes and phase changes—thermal energy rises during heating, while phase energy remains constant during changes.
Calculation Guidance
Determine if the curve involves slanted (temperature change) or flat (phase change) segments.
For slanted lines use Q = mcΔT; for flat lines use Q = mΔH.
Ensure units are appropriate: mass in grams, temperature in Celsius.
If the substance is below zero (solid), between 0 and 100 (liquid), and above 100 (gas).
Specific Heats of Water
Ice: 2.06 J/g°C
Liquid Water: 4.184 J/g°C
Steam: 2.02 J/g°C
When using specific heat values, remember to match them with the state of water.
Phase Change Energies
Heat of Fusion (solid-liquid): 334 J/g
Heat of Vaporization (liquid-gas): 2260 J/g
Determine whether energy is absorbed (positive value) or released (negative value) during phase changes.
Example Problems
One-Step Heating
Example: Heating from 25°C to 77°C.
Use Q = mcΔT with values:
m = 75g, C = 4.184 J/g°C, ΔT = 52°C.
Final energy calculation = 16,000 J (rounded for significant figures).
Phase Change Example
Example: Cooling steam (225g) to water.
Use Q = mΔH_sub_v = 225g * (-2260 J/g) = -508,500 J.
Final answer: -509,000 J (rounded for significant figures).
Multi-Step Problem
Example: Melting ice at 0°C then heating to 23°C.
Calculate total energy for melting then for heating liquid:
Q_fusion = 334 J/g * mass.
Q_heating = mcΔT (with liquid specific heat).
Final calculation: sum energies for each step to obtain total joules.
Final Notes
Practice identifying the type of phase for each step and select the appropriate constants for calculations.
Review the videos and update your summary based on these discussions.