Calorimetry and Thermochemistry Lecture Notes

Key Concepts

• Calorimetry Formula: Q = mCΔT
  • Q: Heat (in kilojoules or joules)
  • m: Mass (in grams)
  • C (or Cs): Specific heat capacity (joules per gram Celsius)
  • ΔT: Change in temperature (final - initial, in Celsius)
• Ensure all values are in specific units before using the formula.

Resources

• 50-page guide on thermochemistry available with common test questions and solutions.
• Find the guide through the link provided in the description box.

Example Problems

Example 1: Finding Heat Released

• Problem: Specific heat capacity of lead = 0.129 J/g°C. Calculate heat released when 497g of lead cools from 37.2°C to 22.5°C.
• Given:
  • Specific heat capacity (C): 0.129 J/g°C
  • Mass (m): 497g
  • Initial temperature: 37.2°C
  • Final temperature: 22.5°C
• Solution Steps:
  1. Calculate ΔT: 22.5°C - 37.2°C = -14.7°C
  2. Use formula: Q = mCΔT
     • Calculate Q: 497g * 0.129 J/g°C * -14.7°C = -942 J
  3. Answer: -942 J or "942 J are released"

Example 2: Finding Specific Heat Capacity

• Problem: 120g aluminum absorbs 9,612 J; temperature increases from 25°C to 115°C. Find specific heat capacity.
• Given:
  • Mass (m): 120g
  • Heat (Q): 9,612 J
  • Initial temperature: 25°C
  • Final temperature: 115°C
• Solution Steps:
  1. Calculate ΔT: 115°C - 25°C = 90°C
  2. Use formula: Q = mCΔT
     • Solve for C: C = Q / (mΔT)
     • Calculate C: 9,612 J / (120g * 90°C) = 0.89 J/g°C
  3. Answer: Specific heat capacity = 0.89 J/g°C

Additional Resources

• Thermochemistry guide for further reading and help.
• Description box includes links for homework help and online tutoring.