Heat Capacity and Specific Heat

Heat Capacity

• Definition: The heat required to raise the temperature of a quantity of substance by one degree Celsius (or one Kelvin).
• Symbol: Capital C.
• Units:
  • Joules per degree Celsius (J/°C)
  • Joules per Kelvin (J/K)
• Explanation:
  • No conversion required between Joules per °C and Joules per K as both represent a unit of temperature change.
• Extensive Property: Depends on the amount of substance.
  • Example: It takes more energy to change the temperature of 1000 mL of water by 1 degree than 100 mL.
• Equation:
  • C = Heat absorbed / Temperature change (ΔT)

Specific Heat (Specific Heat Capacity)

• Definition: The amount of heat required to raise the temperature of one gram of substance by 1 degree Celsius (or 1 Kelvin).
• Symbol: Lowercase c.
• Units:
  • Joules over grams times degrees Celsius (J/g·°C)
  • Joules over grams times Kelvin (J/g·K)
  • Sometimes: Calories over grams per degree Celsius (cal/g·°C)
• Intensive Property: Does not depend on the amount of substance.
• Molar Heat Capacity: Amount of heat required to raise the temperature of one mole of substance by 1 degree Celsius (or 1 Kelvin).
  • Units:
    • Joules over mole degrees Celsius (J/mol·°C)
    • Joules over mole Kelvin (J/mol·K)

Important Equations

• Heat Absorbed or Released:
  • Q = m·c·ΔT
  • Variables:
    • Q: Heat (Joules)
    • m: Mass (grams)
    • c: Specific heat (J/g·°C)
    • ΔT: Change in temperature (°C) = Final temperature - Initial temperature

Example Problem: Calculating Specific Heat

• Problem: Calculate the specific heat of water if 83.68 Joules of heat is required to raise the temperature of 4 grams of water by 5 degrees Celsius.
• Solution:
  • Given: Q = 83.68 J, m = 4.000 g, ΔT = 5.00 °C
  • Equation: Q = m·c·ΔT
  • Rearrange: c = Q / (m·ΔT)
  • Solve: c = 83.68 J / (4.000 g * 5.00 °C) = 0.184 J/g·°C
  • Note: The specific heat of water is 4.184 J/g·°C (memorize this value).*

Specific Heat Values of Common Substances

• Water:
  • Liquid: 4.184 J/g·°C
  • Solid: 2.089 J/g·°C
  • Gas: 2.042 J/g·°C
• Nitrogen Gas: 1.04 J/g·°C
• Metals:
  • Aluminum: 0.892 J/g·°C
  • Gold: 0.129 J/g·°C
  • Iron: 0.442 J/g·°C

Practice Problem: Final Temperature Calculation

• Problem: Calculate the final temperature after 1485 Joules of energy is added to 16.7 grams of water starting at 23.4 °C.

• Solution:

  • Given: Q = 1485 J, m = 16.7 g, T_initial = 23.4 °C, c (water) = 4.184 J/g·°C
  • Equation: Q = m·c·ΔT
  • Find ΔT: ΔT = Q / (m·c) = 1485 J / (16.7 g * 4.184 J/g·°C) = 21.31 °C
  • T_final = ΔT + T_initial = 21.31 °C + 23.4 °C = 44.7 °C.
  • Answer: Final temperature = 44.7 °C.

• Tips for Solving Problems:

  • Identify and write down all variables.
  • Ensure all units match (e.g., grams, joules, Celsius).
  • Write and rearrange the equation as necessary.
  • Substitute values correctly and solve.
  • For temperatures, remember ΔT = Final temperature - Initial temperature.*