# Explain how real machines can never recover the full amount of energy they were supplied with?.

Nov 23, 2016

Sounds like efficiency to me. In general, efficiency is defined as the work output acquired from some work input.

where:

• $w$ is work.
• ${q}_{H}$ is the heat flow into the engine from the hot reservoir.
• ${q}_{C}$ is the heat flow from the engine into the cold reservoir.
• ${T}_{H}$ is the temperature of the hot reservoir.
• ${T}_{C}$ is the temperature of the cold reservoir.

The ideal machine can achieve 100% efficiency, and the work it performs is 100% reversible. Real machines, however, can never quite achieve 100% efficiency, and the work they do is nearly 100% reversible.

As I derived above, a cyclic process has a change in entropy of $\Delta S = 0$, and if heat flow $q$ is reversible, then $\Delta S = {q}_{\text{rev}} / T$.

If we separate the cyclic process into two processes, then

${q}_{H} / {T}_{H} + {q}_{C} / {T}_{C} = 0$

Also from the above, I calculated that

$\textcolor{b l u e}{e} = | w \frac{|}{q} _ H = \frac{{q}_{H} + {q}_{C}}{q} _ H = \textcolor{b l u e}{1 + {q}_{C} / {q}_{H}}$.

${q}_{C}$ flows out of the engine, and thus, ${q}_{C} \le 0$. When ${q}_{C} = 0$, the efficiency is 100%.

Or, using the first relation:

$\textcolor{b l u e}{e} = | w \frac{|}{q} _ H = \frac{{q}_{H} + {q}_{C}}{q} _ H = \textcolor{b l u e}{1 - {T}_{C} / {T}_{H}}$.

So, we could even say that since all absolute temperatures are positive (i.e. when in $\text{K}$), as they are in this equation, $0 \le e \le 1$. When ${T}_{C} = {T}_{H}$, $e = 0$, and when ${T}_{C} = \text{0 K}$ (which has yet to be accomplished!), or ${T}_{H}$ is absurdly large, $e \approx 1$.

Therefore, the efficiency can never be more than 100%, and for real machines, the efficiency is effectively never exactly 100%.