Problem — An engine with efficiency greater than 1

Problem

Imagine a device that produces more work than the heat it absorbs from the hot source ($\eta > 1$). Which law of thermodynamics would it violate? And if instead $\eta = 1$ exactly (with no heat rejected to the cold side), which law would it violate?

Solution

Definition. The efficiency is $\eta = L/Q_c$, with $L$ the work produced and $Q_c$ the heat absorbed from the hot source.

Case $\eta > 1$. This would mean $L > Q_c$: the engine would produce more energy (as work) than it absorbs as heat. Over a cycle the internal energy returns to its initial value ($\Delta U = 0$), so the first law requires

$L = Q_c - |Q_f| \le Q_c$

Having $L > Q_c$ would mean creating energy out of nothing: it violates the first law (conservation of energy). This is the classic perpetual motion machine of the first kind.

Case $\eta = 1$. Here $L = Q_c$ and $|Q_f| = 0$: all the absorbed heat becomes work, with nothing given up to the cold reservoir. Energy is conserved (the first law holds), but such a device — which draws heat from a single source and converts it entirely into work in a cyclic process — violates the second law in the Kelvin statement. This is the perpetual motion machine of the second kind.

$\ev{\eta > 1 \Rightarrow \text{violates the 1st law}; \quad \eta = 1 \Rightarrow \text{violates the 2nd (Kelvin)}}$

Every real heat engine therefore has $\eta < 1$ strictly, and is further bounded by the Carnot value $\eta_\text{max} = 1 - T_f/T_c$.

Collegamenti

Argomenti: Macchine termiche Concetti: Rendimento · Primo principio della termodinamica · Secondo principio della termodinamica Competenze: Analisi di casi limite e fantafisica