ΔS of a mixture at different temperatures

Problem

$0{,}5\;\text{kg}$ of water at $80\;\text{°C}$ is mixed with $0{,}5\;\text{kg}$ of water at $20\;\text{°C}$. Calculate (a) the equilibrium temperature, (b) the total $\Delta S$ (specific heat $c=4186\;\text{J/(kg·K)}$).

Solution

(a) The masses are equal, so the equilibrium temperature is the arithmetic mean of the absolute temperatures $T_1=353\;\text{K}$ and $T_2=293\;\text{K}$: $T_e=\frac{353+293}{2}=323\;\text{K}\quad(50\;\text{°C})$ (b) The hot water cools from $353\;\text{K}$ to $323\;\text{K}$: $\Delta S_\text{hot}=mc\ln\frac{T_e}{T_1}=0{,}5\cdot 4186\cdot\ln\frac{323}{353}=2093\cdot(-0{,}0888)=-185{,}9\;\text{J/K}$ The cold water warms from $293\;\text{K}$ to $323\;\text{K}$: $\Delta S_\text{cold}=mc\ln\frac{T_e}{T_2}=0{,}5\cdot 4186\cdot\ln\frac{323}{293}=2093\cdot(+0{,}0975)=+204{,}1\;\text{J/K}$ Summing: $\Delta S_\text{tot}=-185{,}9+204{,}1\approx +17{,}8\;\text{J/K}>0$ The result is positive, as required by the second law for an irreversible process. $\ev{T_e=323\;\text{K},\ \Delta S\approx 17{,}8\;\text{J/K}}$

Links

Topics: Entropy and the second law Concepts: Entropy · Temperature · Second law of thermodynamics Skills: Entropy balance · Calorimetric equation