Worked exercise — maximum work from cooling coffee

Problem

The same coffee as before ($0{,}1\;\text{kg}$, from $353\;\text{K}$ to $293\;\text{K}$), but instead of cooling directly, it passes through a heat engine that extracts work $L$ and releases $Q_\text{atm}$ to the atmosphere. Find the maximum work obtainable.

Solution

Setup. The maximum work is obtained when the process is reversible, i.e. when $\Delta S_\text{tot} = 0$. There are three components: the coffee (which cools), the engine (which performs a cycle) and the atmosphere (a thermostat at $293\;\text{K}$).

The engine works cyclically, so it returns each time to the same state: $\Delta S_\text{engine} = 0$. We have already calculated the entropy of the cooling coffee: $\Delta S_\text{coffee} = -77{,}9\;\text{J/K}$.

Reversibility condition ($\Delta S_\text{tot} = 0$): $\underbrace{-77{,}9}_\text{coffee} + \underbrace{0}_\text{engine} + \frac{Q_\text{atm}}{293} = 0 \;\Rightarrow\; Q_\text{atm} = 77{,}9\cdot 293 = 22\,824\;\text{J}$

First law. The heat leaving the coffee ($25\,080\;\text{J}$, as in the direct-cooling exercise) is split between the heat released to the atmosphere and the useful work extracted: $L_\text{max} = 25\,080 - 22\,824 = \ev{2256\;\text{J}}$

This is the maximum work extractable from a cup of coffee cooling from $80\;^\circ\text{C}$ to room temperature. Any real engine will have $\Delta S_\text{tot} > 0$ and will produce less work: irreversibility is, literally, wasted work.

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Topics: Entropia e secondo principio Concepts: Secondo principio della termodinamica · Entropia · Rendimento Skills: Bilancio entropico Objects: Macchina termica