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The science
Fourier conduction on a lattice
Cell i exchanges heat with its neighbours: dT_i/dt = k (T_(i-1) - 2 T_i + T_(i+1)), with the two end cells clamped to the reservoir temperatures T0 + dT/2 and T0 - dT/2. The flux across each link is J_i = k (T_i - T_(i+1)).
Steady state
The profile relaxes to a straight line and every link carries the same flux J = k dT / (N - 1) for ever: a non-equilibrium steady state. The slowest relaxation mode decays at rate lambda = 2k (1 - cos(pi / (N - 1))), about k pi^2 / (N-1)^2 for a long chain, so a longer chain takes quadratically longer to settle.
Entropy production
Each link produces sigma_i = J_i (1/T_(i+1) - 1/T_i) >= 0. In the steady state the sum telescopes: S_dot = J (1/T_cold - 1/T_hot). It is zero only at dT = 0 (equilibrium), and positive for any flow: the second law in action.
Honest simplifications: one-dimensional chain, constant conductivity, no heat capacity differences, reservoirs are ideal, the cells in the 3D picture across the width share one temperature. The time unit is the model's own, not seconds.
What to try
- Press Restart and watch the flat chain relax to a straight line; compare the decay with the relaxation time in the readouts.
- Double the chain length and see the steady-state flux halve and the relaxation take about four times longer.
- Switch the gradient off: flux and entropy production fall to zero.
- Raise the mean temperature: the same flux produces less entropy because S_dot scales as 1/T squared.