Worked exercise — the coffee cools down

Problem

$0{,}1\;\text{kg}$ of coffee (approximated as water, $c_s = 4180\;\text{J/(kg·K)}$) at $T_A = 80\;^\circ\text{C} = 353\;\text{K}$ is left to cool down to $T_B = 20\;^\circ\text{C} = 293\;\text{K}$ in a room at $20\;^\circ\text{C}$ (thermostat). Calculate $\Delta S_\text{tot}$.

Solution

Setup. The isolated system comprises two components: the coffee (which cools) and the room’s atmosphere, which behaves as a thermostat at $T_\text{atm} = 293\;\text{K}$. We compute the two entropy changes separately and add them.

Entropy of the coffee (decreases, since it cools): $\Delta S_\text{caffè} = m c_s \ln\frac{T_B}{T_A} = 0{,}1\cdot 4180\cdot\ln\frac{293}{353} = -77{,}9\;\text{J/K}$

Heat given to the thermostat. The heat leaving the coffee enters the atmosphere: $Q = m c_s (T_A - T_B) = 0{,}1\cdot 4180\cdot 60 = 25\,080\;\text{J}$

Entropy of the thermostat (the atmosphere receives $Q$ at constant temperature): $\Delta S_\text{atm} = \frac{Q}{T_\text{atm}} = \frac{25\,080}{293} = 85{,}6\;\text{J/K}$

Total change: $\Delta S_\text{tot} = -77{,}9 + 85{,}6 = \ev{+7{,}7\;\text{J/K} > 0}$

The result is strictly positive, so the process is irreversible: the coffee would never spontaneously turn hot again, even though total energy would be conserved. The decrease in the coffee’s entropy is more than compensated by the increase in the environment’s entropy.

What if we heated it up again?

If we tried to reheat the coffee by putting it back in an oven at $80\;^\circ\text{C}$, the total entropy would grow again. The coffee’s entropy would return to its initial value ($+77{,}9\;\text{J/K}$), but the oven’s would decrease only by $25\,080/353 = 71{,}0\;\text{J/K}$. The difference $+6{,}9\;\text{J/K}$ is the irreversibility of the whole cycle: the universe never returns to its initial state.

Collegamenti

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