Energy Basics

1. The Nature of Energy

Definition: The capacity to supply heat or do work.

  • Potential Energy (PE): Energy of position or composition (e.g., chemical bonds in glucose).
  • Kinetic Energy (KE): Energy of motion (e.g., blood flowing through an artery).

Law of Conservation of Energy: Energy is neither created nor destroyed, only transformed.
Clinical Note: In the body, chemical PE in food is converted to KE (muscle contraction) and Thermal Energy (body heat).

2. Thermal Energy, Temperature, and Heat

These three terms are often confused but distinct:

  • Thermal Energy: Total KE of random motion of atoms/molecules.
  • Temperature: The average KE of particles (the "intensity" of heat).
  • Heat (q): The transfer of thermal energy between two bodies at different temperatures.
Direction of Flow: Heat always moves from High T → Low T until thermal equilibrium is reached.

3. Exothermic vs. Endothermic

Process Energy Change Allied Health Example
Exothermic Releases heat to surroundings Metabolic breakdown of sugar; "Hot packs" for stiffness.
Endothermic Absorbs heat from surroundings Evaporation of sweat (cools the skin); Instant "Cold packs" (Ammonium Nitrate).

4. Units of Energy

  • Joule (J): The SI unit.
  • calorie (cal): Energy to raise 1g of water by 1°C.
  • Calorie (C): 1 kcal or 1,000 calories (Used in nutrition).

1 calorie = 4.184 Joules

5. Heat Capacity vs. Specific Heat

Heat Capacity (C): Extensive property (depends on how much matter you have).
C = q / ΔT

Specific Heat (c): Intensive property (depends only on the identity of the substance).
c = q / (m × ΔT)

6. Calculating Heat Transfer

To find the total heat gained or lost by a system:

q = m × c × ΔT

Where:
m = mass (g)
c = specific heat (J/g°C)
ΔT = Tfinal - Tinitial

Practice Problem 1 (The Fever):
A patient with a high fever is cooled using 500g of water in a sponge bath. If the water temperature rises from 20°C to 30°C, how much heat (in Joules) did the water absorb? (Specific heat of water = 4.184 J/g°C).

Solution: q = (500g)(4.184 J/g°C)(10°C) = 20,920 J
Practice Problem 2 (Material Comparison):
If you apply 100 J of heat to 10g of Gold (c = 0.129 J/g°C) and 10g of Aluminum (c = 0.897 J/g°C), which one will experience a higher temperature change?

Answer: Gold. Because it has a much lower specific heat, it takes less energy to change its temperature.