Theory: Kelvin–Planck Statement (Second Law)
1) The Kelvin–Planck statement
The Kelvin–Planck statement of the second law says:
a device operating in a cycle cannot take heat from a single thermal reservoir and convert it completely into work.
Real heat engines must reject some heat to a colder sink (\(Q_c>0\)).
That “waste” heat is not a design flaw — it’s a fundamental second-law constraint.
2) First law vs. second law
For a cyclic engine (no net internal energy change), the first-law balance in magnitude form is:
The first law alone does not forbid \(Q_c=0\); it simply permits \(W=Q_h\).
The second law is what forbids 100% conversion from a single reservoir.
3) Entropy proof (why \(\eta=1\) fails)
Consider a hot reservoir at temperature \(T_h\). If the engine absorbs heat \(Q_h\) from it, the reservoir’s entropy change is:
If the engine rejects heat \(Q_c\) to a cold reservoir at \(T_c\), the cold reservoir gains entropy:
Total entropy change of the universe (reservoirs + device over a cycle):
If \(Q_c=0\) (i.e., \(\eta=1\)), then
\(\Delta S_{univ} = -Q_h/T_h < 0\), which violates the second law.
4) Carnot limit and why \(\eta\le\eta_{Carnot}\)
For an engine operating between two reservoirs at \(T_h\) and \(T_c\), the maximum possible efficiency is the Carnot efficiency:
Any claim with \(\eta>\eta_{Carnot}\) implies \(\Delta S_{univ}<0\) for the corresponding heat transfers, so it is impossible.
5) Equivalence with the Clausius statement (the “combo” demo)
The Kelvin–Planck and Clausius statements are equivalent: if one could be violated, the other could be violated too.
A standard argument is:
- Assume a 100% engine exists (\(\eta=1\), so \(W=Q_h\), \(Q_c=0\)).
- Use the work output to power a refrigerator, moving heat from cold to hot.
- Because the work came “for free” from a single reservoir, the combined device can transfer heat cold \(\to\) hot with no external work — a Clausius violation.
6) Practical notes (real engines)
- Real engines have additional irreversibilities, so their efficiencies are usually well below Carnot.
- Reversible (Carnot) is an ideal benchmark: \(\Delta S_{univ}=0\).
- Irreversible engines: \(\Delta S_{univ}>0\) and \(\eta<\eta_{Carnot}\).
