Heating of water: ΔS

Problem

A mass of $1{,}0\ \text{kg}$ of water is heated reversibly from $20\ ^\circ\text{C}$ to $80\ ^\circ\text{C}$. Calculate the entropy change (specific heat $c=4186\ \text{J/(kg·K)}$).

Solution

Reversible heating: the infinitesimal heat exchange is $\dd Q=mc\,\dd T$, so the entropy change is obtained by integrating $\dd S=\dfrac{\dd Q}{T}$. Converting the temperatures to Kelvin: $T_A=20+273=293\ \text{K}$, $T_B=80+273=353\ \text{K}$. $\Delta S=\int_{T_A}^{T_B}\frac{mc\,\dd T}{T}=mc\ln\frac{T_B}{T_A}=1{,}0\cdot 4186\cdot\ln\frac{353}{293}\approx 780\;\text{J/K}$ The positive sign is expected: heating the water increases the number of accessible microstates for the molecules. $\ev{\Delta S\approx +780\;\text{J/K}}$

Collegamenti

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