Entropy and the second law: what forbids

The first law only balances

• The first law is an accountant: it balances the books on any cycle, however bad. The second law is a judge: some balanced books describe machines that cannot exist - and entropy 熵 is the quantity that says which.
• $ds = \delta q_{rev}/T$ defines entropy from a reversible path; the entropy balance is $\sigma \ge 0$ for every process, zero only for a reversible process 可逆过程. On a T-s diagram the reversible heat is the area under the curve.

The Carnot bound

• The Carnot cycle 卡诺循环 - two isotherms, two adiabats, all reversible - is the best cycle between two reservoirs, and its thermal efficiency 热效率 is $\eta_C = 1 - T_L/T_H$: between $800\ \mathrm{K}$ and $300\ \mathrm{K}$, $\eta_C = \mathbf{0.625}$. A claimed 0.70 engine between those reservoirs is impossible on paper, no prototype needed.
• The Kelvin-Planck statement is the reason: no cycle converts heat from one reservoir fully into work. Any claim that beats Carnot violates it, and reversible operation is the ceiling because running a better-than-reversible engine backwards would break the law outright.

The irreversible valve

• Throttling 节流 keeps enthalpy, not entropy: steam at $1000\ \mathrm{kPa}$ ($h = 3059\ \mathrm{kJ/kg}$, $s = 7.155$) throttled to $200\ \mathrm{kPa}$ lands at the same $h$ with $s_2 = 7.289$, so $\sigma = \mathbf{0.134}\ \mathrm{kJ/(kg\,K)} > 0$ - irreversible with no heat transfer at all.
• The entropy balance finds irreversibility without needing to see the mess: $\sigma$ is the accounting quantity of the forbidden.

The Carnot ledger and lost work

• A Carnot engine takes $1000\ \mathrm{kJ}$ from $800\ \mathrm{K}$ and rejects to $300\ \mathrm{K}$: $W = \mathbf{625}\ \mathrm{kJ}$, $Q_L = \mathbf{375}\ \mathrm{kJ}$, $\Delta S_{hot} = -\mathbf{1.250}$ and $\Delta S_{cold} = +\mathbf{1.250}\ \mathrm{kJ/K}$ - the universe is unchanged, which is what reversible means.
• Burn the same fuel irreversibly at $2000\ \mathrm{K}$ before transferring the heat: $\sigma = Q(1/800 - 1/2000) = \mathbf{0.75}\ \mathrm{kJ/K}$, and the lost work 损失功 $T_0 \sigma = 300 \times 0.75 = \mathbf{225}\ \mathrm{kJ}$ - exactly the exergy destroyed.

Three second-law slips

Recap