What you'll be able to do本章學習成果
- State the second law three ways — Clausius, Kelvin–Planck, and entropy — and explain internally reversible processes and the Kelvin temperature scale.
- List the common irreversibilities of engineering practice.
- Assess power, refrigeration, and heat-pump cycles against their reversible (Carnot) limits.
- Interpret the Clausius inequality to decide whether a cycle is reversible, irreversible, or impossible.
Key equations重要公式
Why a second law?為何需要第二定律？
Conservation of mass and energy tell you the disposition of mass and energy in a process — but not whether the process can actually happen. A cup of coffee never spontaneously reheats itself by drawing energy from the cooler room, even though energy would be perfectly conserved if it did. The second law is the guiding principle for which direction processes run. It also lets us establish equilibrium conditions, define an absolute temperature scale, and pin down the best theoretical performance any device can reach.
Three statements, one law
No single sentence captures the whole second law; three classic formulations are equivalent:
It is impossible for any system to operate so that the sole result is energy transfer by heat from a cooler to a hotter body. (Heat won't flow uphill on its own.)
It is impossible for any system to operate in a cycle and deliver net work to its surroundings while exchanging heat with only a single thermal reservoir. (You can't build a cycle that just turns heat from one reservoir entirely into work.)
It is impossible for any system to operate in a way that entropy is destroyed. Unlike mass and energy, entropy is produced — never consumed — whenever irreversibilities are present. (Developed fully in the Entropy chapter.)
Thermal reservoirs熱力學熱庫
A thermal reservoir stays at constant temperature no matter how much energy is added or removed by heat transfer. The atmosphere, oceans and lakes, or a large block of copper all approximate one. Reservoirs give us fixed hot and cold temperatures to analyze cycles against.
Irreversibilities不可逆性
Actual processes differ from idealized ones through irreversibilities. The common ones in engineering:
- Heat transfer across a finite temperature difference
- Unrestrained (free) expansion to lower pressure
- Spontaneous chemical reaction and mixing
- Friction — sliding and fluid
- Electric current through a resistance; hysteresis; inelastic deformation
A process is reversible if no irreversibilities exist in either the system or surroundings (a full idealization). It is internally reversible if none exist within the system — a quasi-equilibrium process — even if external ones (like heat transfer across a gap) do. All real processes are irreversible.
Kelvin–Planck, quantified克耳文–蒲朗克陳述的量化
For any system undergoing a cycle while exchanging heat with a single reservoir, the net work can only be negative or zero — never positive:
For a power cycle between two reservoirs, efficiency is the fraction of $Q_H$ converted to work:
The law forces $\eta < 100\%$: some heat must always be discharged to the cold reservoir.
Carnot corollaries卡諾推論
Two deductions from Kelvin–Planck shape everything that follows:
- An irreversible power cycle always has lower efficiency than a reversible one operating between the same two reservoirs.
- All reversible cycles between the same two reservoirs have the same efficiency — independent of working fluid or mechanism.
The same logic applies to refrigeration and heat-pump cycles in terms of their coefficients of performance.
Maximum performance最大性能
Because reversible cycles set the ceiling, and (by the Kelvin scale) the reversible heat-transfer ratio equals the temperature ratio $Q_C/Q_H = T_C/T_H$, the maximum theoretical measures depend only on reservoir temperatures (in Kelvin):
These are the Carnot limits for the power, refrigeration, and heat-pump cycles respectively. No real device beats them.
Carnot & Clausius explorer卡諾與克勞修斯探索器
Set the reservoir temperatures and energy transfers for a power cycle, refrigerator, or heat pump. The schematic scales its arrows to the energies, and the verdict tells you whether your numbers describe a reversible, irreversible, or impossible cycle — by comparing to the Carnot limit and computing the Clausius cyclic integral $\sigma_{cycle}$.
Try TH = 500 K, TC = 300 K, Q_H = 1000 kJ. Q_C = 600 kJ is exactly reversible; Q_C = 400 kJ is impossible; Q_C = 700 kJ is irreversible.
The Carnot cycle卡諾循環
The Carnot cycle is a concrete reversible cycle between two reservoirs: four internally reversible processes — two isothermal alternating with two adiabatic. Run forward it is a power cycle with $\eta = 1 - T_C/T_H$; run backward, the same energy transfers reverse direction, giving a Carnot refrigerator or heat pump with the COP limits above. (The Stirling and Ericsson cycles are other reversible benchmarks.)
The Clausius inequality克勞修斯不等式
Applicable to any cycle, the Clausius inequality is the bridge to entropy:
where the integral runs over the whole boundary and the whole cycle. The sign of $\sigma_{cycle}$ classifies the cycle:
- $\sigma_{cycle} = 0$ — no internal irreversibilities (reversible)
- $\sigma_{cycle} > 0$ — irreversibilities present
- $\sigma_{cycle} < 0$ — impossible
The explorer above computes exactly this quantity for the cycle you dial in.
Is the cycle possible?此循環是否可能存在？
Example範例 Power cycle vs. the Carnot limit動力循環與卡諾極限比較 ›
Given: a power cycle receives 1000 kJ at $T_H=500$ K and rejects 600 kJ at $T_C=300$ K.
Find: whether it is reversible, irreversible, or impossible.
Solution. Actual $\eta=1-600/1000=0.40$. Carnot limit $\eta_{max}=1-300/500=0.40$. Since $\eta=\eta_{max}$, the cycle is reversible. Check Clausius: $\sigma=\tfrac{600}{300}-\tfrac{1000}{500}=0$. ✓