Entropy and the second law: what forbids
Introduzido| English | Significado |
|---|---|
| entropy/ˈentrəpi/ | A thermodynamic measure related to energy dispersal and the number of microscopic arrangements consistent with a system's macroscopic state. |
| reversible process | reversible process |
| Carnot cycle | Carnot cycle |
| thermal efficiency | thermal efficiency |
| throttling | throttling |
| lost work | lost work |
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.
ds = delta q / T may be applied directly when:
Dificuldade 1/5 · Moedas 100 + pequeno bônus na primeira conclusão hoje
The definition needs q_rev; sigma catches the rest.
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.
Reversible, irreversible, or bound?
Sort each statement by its second-law role.
Carnot efficiency between 800 K and 300 K (three decimals)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
1 - 300/800.
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.
Throttling sigma, 1000 to 200 kPa (three decimals, kJ/(kg K))?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
7.289 - 7.155.
Throttling keeps entropy constant.
Dificuldade 1/5 · Moedas 100 + pequeno bônus na primeira conclusão hoje
It keeps enthalpy; sigma = 0.134 kJ/(kg K) at these states.
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.
Carnot work from 1000 kJ at 800 K rejecting to 300 K (kJ)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
0.625 x 1000.
Entropy change of each reservoir (three decimals, kJ/K)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
1000/800 = 375/300.
sigma for burning-transfer at 2000 K then 800 K (two decimals, kJ/K)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
1000 x (1/800 - 1/2000).
Lost work at T0 = 300 K (kJ)?
Dificuldade 2/5 · Moedas 140 + pequeno bônus na primeira conclusão hoje
T0 sigma.
Three second-law slips
- Applying $ds = \delta q/T$ to an irreversible path. It needs $q_{rev}$ - or use the entropy balance with $\sigma$.
- Reading an efficiency claim without its reservoir temperatures. The bound is $1 - T_L/T_H$ in absolute kelvin.
- Assuming a throttle keeps $s$. It keeps $h$; $\sigma = 0.134$ says otherwise.
Recap
- $\sigma \ge 0$ for everything; $ds = \delta q_{rev}/T$; T-s area is reversible heat.
- $\eta_C = 1 - T_L/T_H = 0.625$ between 800 K and 300 K - the paper veto on bad machines.
- Lost work $= T_0 \sigma$: the 225 kJ that irreversible burning never delivers.