Cooled iron: maximum work

Problem

A block of iron ($m=1$ kg, specific heat $c_s = 450\ \text{J/(kg}\cdot\text{K)}$) at temperature $T_A = 500$ K is cooled down to room temperature $T_0 = 300$ K. Calculate the maximum work extractable by inserting a reversible cyclic machine between the iron and the environment. Compare with the total thermal energy given up $Q_\text{tot}$ and comment.

Solution

The extracted work is maximum when the machine operates reversibly, that is, when total entropy does not change: $\Delta S_\text{tot}=0$.

The iron cools from $500$ K to $300$ K; its entropy change (variable temperature) is $\Delta S_\text{ferro} = m c_s\ln\frac{T_0}{T_A} = 1\cdot 450\cdot\ln\frac{300}{500} = 450\cdot\ln 0{,}6 = -229{,}9\;\text{J/K}$

Reversibility condition: the environment’s entropy exactly compensates that lost by the iron, $\Delta S_\text{amb} = +229{,}9\ \text{J/K}$. The environment is a thermostat at $T_0$, so it receives a heat $Q_\text{amb} = T_0\,\Delta S_\text{amb} = 300\cdot 229{,}9 = 68\,970\;\text{J}$

Total thermal energy given up by the iron: $Q_\text{tot} = m c_s (T_A - T_0) = 1\cdot 450\cdot 200 = 90\,000\;\text{J}$

The maximum work is the part of $Q_\text{tot}$ not discharged to the environment: $L_\text{max} = Q_\text{tot} - Q_\text{amb} = 90\,000 - 68\,970 \approx 2{,}1\cdot 10^4\;\text{J}$ $\ev{L_\text{max} \approx 2{,}1\cdot 10^4\;\text{J}}$ Comment: only about $23\%$ of the thermal energy given up by the iron can be converted into work; the rest ($\approx 69$ kJ) inevitably ends up in the environment. The useful fraction is limited by the second law, because the iron gives up heat at temperatures that decrease progressively towards $T_0$.

Collegamenti

Argomenti: Entropy and the second law Concetti: Entropy · Second law of thermodynamics · Specific heat and heat capacity · Efficiency Competenze: Entropy balance Oggetti: Heat engine