Specific Heat Notes

Specific Heat

  • Specific heat is the amount of energy needed to raise the temperature of one gram of a substance by one degree Celsius.
  • Relates joules and calories.

Joules and Calories

  • 1 calorie = 4.184.18 joules
  • This value matches the specific heat of water.
  • Important because a large portion of our bodies is water.

Units for Specific Heat

  • Joules per gram per degree Celsius: Jg⋅°C\frac{J}{g \cdot °C}
  • Symbol for specific heat: C (though some use S).
  • C = energy per gram degree Celsius.
  • Joules = energy, grams = mass, degrees Celsius = temperature.

Equation

  • Energy is symbolized by Q (heat).
  • Mass is m.
  • Change in temperature is ΔT\Delta T.
  • Equation: q=mcΔTq = mc \Delta T

Example Problem 1: Evaporative Cooling

  • Calculate heat removed from 2121 grams of water when it cools from 3434 to 2828 degrees Celsius.
  • Givens:
    • Q = unknown (heat in joules)
    • m = 2121 grams
    • Tinitial=34°CT_{initial} = 34°C
    • Tfinal=28°CT_{final} = 28°C
  • Equation: q=mcΔTq = mc \Delta T
    • Specific heat of water (C) = 4.18Jg⋅°C4.18 \frac{J}{g \cdot °C}
  • Calculation:
    • q=(21 grams)⋅(4.18Jg⋅°C)⋅(28°C−34°C)q = (21 \text{ grams}) \cdot (4.18 \frac{J}{g \cdot °C}) \cdot (28°C - 34°C)
    • q=−526.68q = -526.68 joules
    • q≈−527q \approx -527 joules (with sig figs)
  • Negative sign indicates energy is being lost from the system.

Example Problem 2

  • Calculate the specific heat of 2222 grams of a metal that absorbs 713713 joules when heated from 28.328.3 to 72.872.8 degrees Celsius.
  • Givens:
    • m = 2222 grams
    • q = 713713 joules (positive since energy is absorbed)
    • Tinitial=28.3°CT_{initial} = 28.3°C
    • Tfinal=72.8°CT_{final} = 72.8°C
    • C = unknown
  • Equation: q=mcΔTq = mc \Delta T
  • Calculation:
    • 713=22⋅c⋅(72.8−28.3)713 = 22 \cdot c \cdot (72.8 - 28.3)
    • 713=22⋅c⋅44.5713 = 22 \cdot c \cdot 44.5
    • 713=979⋅c713 = 979 \cdot c
    • c=713979=0.728c = \frac{713}{979} = 0.728 joules per gram degree Celsius.
    • c≈0.729Jg⋅°Cc \approx 0.729 \frac{J}{g \cdot °C}

Example Problem 3

  • The specific heat of silver is 0.24Jg⋅°C0.24 \frac{J}{g \cdot °C}. If 15.415.4 grams of silver absorbs 332332 joules of energy, how much will the temperature increase?
  • Givens:
    • C = 0.24Jg⋅°C0.24 \frac{J}{g \cdot °C}
    • m = 15.415.4 grams
    • q = 332332 joules
    • ΔT\Delta T = unknown
  • Equation: q=mcΔTq = mc \Delta T
  • Calculation:
    • 332=15.4⋅0.24⋅ΔT332 = 15.4 \cdot 0.24 \cdot \Delta T
    • 332=3.696⋅ΔT332 = 3.696 \cdot \Delta T
    • ΔT=3323.696=89.827\Delta T = \frac{332}{3.696} = 89.827 degrees Celsius
    • ΔT≈90°C\Delta T \approx 90°C