Entropy

humans are temporary pockets of low entropy

the universe - chapter 1 page 42

Classical Thermodynamics

• Entropy Change for a Reversible Process: $$ \Delta S = \int \frac{dq_{rev}}{T} $$
• Entropy of an Ideal Gas: $$ S = nR \ln \frac{VT^{3/2}{n_0V_0T_0} + S_0 $$}

Statistical Mechanics

• Boltzmann's Entropy Formula: $$ S = k_B \ln W $$
• Gibbs Entropy Formula: $$ S = -k_B \sum_i p_i \ln p_i $$

Information Theory

• Shannon Entropy: $$ H(X) = -\sum_i p(x_i) \log_2 p(x_i) $$

turds

Entropy of Human Feces

We can estimate the entropy (( S )) of feces using the Boltzmann entropy formula:

$$ S = k_B \ln(\Omega) $$

Where: - ( S ) = entropy (in Joules per Kelvin) - ( k_B ) = Boltzmann constant (( 1.38 \times 10^{-23} \, \text{J/K} )) - ( \Omega ) = number of accessible microstates for the system

Step-by-Step Calculation:

1. Assume a simplified system:
2. Feces is made up of ( N ) particles (e.g., molecules or atoms).
3. Each particle can occupy ( m ) possible states (e.g., positions, energy levels).

4. The total number of microstates is given by: $$ \Omega = m^N $$

5. Insert values:

6. Let ( N ) = ( 10^{23} ) (approximate number of molecules in 1 kg of feces).
7. Let ( m ) = ( 10^6 ) (approximate number of possible states per molecule, based on chemical diversity).

Substituting into ( \Omega ): $$ \Omega = (10⁶⁾ $$}

1. Compute the entropy: Taking the natural logarithm: $$ \ln(\Omega) = N \ln(m) = 10^{23} \ln(10^6) $$

Simplify: $$ \ln(\Omega) = 10^{23} \cdot 6 \ln(10) = 10^{23} \cdot 13.8155 $$

Substituting back into the entropy formula: $$ S = k_B \cdot \ln(\Omega) = (1.38 \times 10^{-23}) \cdot (10^{23} \cdot 13.8155) $$

Simplify: $$ S = 1.38 \cdot 13.8155 \approx 19.06 \, \text{J/K} $$

Final Result:

The estimated entropy of 1 kg of feces is approximately:

$$ S \approx 19.06 \, \text{J/K} $$

This is a rough estimation, assuming uniform particle distribution and simplified chemical states.