Hot shower: ΔS of the universe

Problem

During a hot shower $2$ kg of water at temperature $T_\text{calda} = 315$ K enter the bathroom; the water cools until it mixes with the room’s air at $T_\text{aria} = 293$ K, which we can treat as a thermostat. Calculate the increase in entropy of the universe (specific heat of water $c_s = 4180\ \text{J/(kg}\cdot\text{K)}$). Is the process reversible?

Solution

The water cools from $315$ K to $293$ K; its entropy (variable temperature) changes by $\Delta S_\text{acqua} = m c_s\ln\frac{T_\text{aria}}{T_\text{calda}} = 2\cdot 4180\cdot\ln\frac{293}{315} = -605\;\text{J/K}$

Heat given by the water to the air: $Q = m c_s (T_\text{calda} - T_\text{aria}) = 2\cdot 4180\cdot 22 = 183\,920\;\text{J}$

The air, a thermostat at $T_\text{aria}$, receives this heat: $\Delta S_\text{aria} = \frac{Q}{T_\text{aria}} = \frac{183\,920}{293} = +628\;\text{J/K}$

Total balance (universe): $\Delta S_\text{tot} = \Delta S_\text{acqua} + \Delta S_\text{aria} = -605 + 628 \approx +22\;\text{J/K}$ $\ev{\Delta S_\text{tot} \approx 22\;\text{J/K}}$ Since $\Delta S_\text{tot} > 0$, the process is IRREVERSIBLE: heat passes spontaneously from the hot water to the colder air across a finite temperature difference.

Collegamenti

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