LEMNISCATA
Matemàtiques
\begin{equation}
-18º C_{gel} \longrightarrow0º C_{gel} \longrightarrow0º C_{aigua} \longrightarrow100º C_{aigua} \longrightarrow100º C_{vapor} \longrightarrow150º C_{vapor}
\end{equation}
Les dades del problema són:
\begin{equation}
\left\{\begin{array}{lllll} &C_{p_{gel}}=0.5\ \text{cal/g · K} \\
&C_{p_{aigua}}=1.0\ \text{cal/g · K} \\
&C_{p_{vapor d’aigua}}=0.47\ \text{cal/g · K}\\
&l_{f_{gel}}=80\ \text{cal/g} \\
&l_{v_{aigua}}=80\ \text{cal/g}
\end{array}\right.
\end{equation}
ara calcularem l’increment d’entropia per a cada procés, tal com:
\begin{equation}
\boxed{\Delta S_1}=\displaystyle\int_{273-18}^{273+0}\frac{C_{gel}}{T}dT=\displaystyle\left. C_{gel}\ln T\displaystyle\right]_{255}^{273}=0.5\ \text{cal/g · K}\cdot\ln\left(\frac{273}{255}\right)=\boxed{0.034\ \text{cal/g · K}}
\end{equation}
\begin{equation}
\boxed{\Delta S_2}=\int_{}^{}\frac{dQ_{rev}}{T}=\frac{l_f}{T}=\frac{80\ \text{cal/g}}{273\ \text{K}}=\boxed{0.293\ \text{cal/g · K}}
\end{equation}
\begin{equation}
\boxed{\Delta S_3}=\int_{273+0}^{273+100}\frac{C_{aigua}}{T}dT=\left. C_{aigua}\ln T\right]_{273}^{373}=1.0\ \text{cal/g · K}\cdot\ln\left(\frac{373}{273}\right)=\boxed{0.312\ \text{cal/g · K}}
\end{equation}
\begin{equation}
\boxed{\Delta S_4}=\int_{}^{}\frac{dQ_{rev}}{T}=\frac{l_v}{T}=\frac{540\ \text{cal/g}}{373\ \text{K}}=\boxed{1.448\ \text{cal/g · K}}
\end{equation}
\begin{equation}
\boxed{\Delta S_5}=\int_{273+100}^{273+150}\frac{C_{vapor}}{T}dT=\left. C_{vapor}\ln T\right]_{373}^{423}=0.47\ \text{cal/g · K}\cdot\ln\left(\frac{423}{373}\right)=\boxed{0.059\ \text{cal/g · K}}
\end{equation}
L’increment d’entropia total serà la suma de cadascuna de les entropies:
\begin{equation}
\boxed{\Delta S_{TOTAL}} =\sum_{i=n}^{n=5}\Delta S_i=\Delta S_1+\Delta S_1+\Delta S_3+\Delta S_4+\Delta S_5=
\end{equation}
\begin{equation}
=0.034+0.293+0.312+1.448+0.059=\boxed{2.146\ \text{cal/g · K}}
\end{equation}