# Thermal Hysteresis in Melting-Solidification of Nanoparticles
## 2 Results and Discussion
### Analysis of NP Solidification  
The solidification processes in the NPs of different sizes can be analyzed within the CNT on the basis of an equal nucleation probability, i.e., \(P(R,T)=Const\). This means that by choosing a reference particle of a radius \(R_{0}\) and a solidification temperature \(T_{s0}\) and equating the nucleation probabilities  
\[P(R,T_{s})=P(R_{0},T_{s0}) \tag{5}\]  
we can express the solidification parameters for the NPs of a radius \(R\) in terms of those for the reference particle. The solidification of the NPs proceeds from the formation of a solid nucleus of a radius \(x\). To implement Equation (2), we require a suitable expression for the nucleation barrier \(\Delta F_{N}(R,T_{s})\). As the surface energy for the solid state is always higher than that for the liquid state, we can assume that the solid nucleus is always surrounded by liquid to minimize the energy of the particle. Assuming (for simplicity) that the nucleus is formed in the NP's center, the excess free energy associated with it takes the form  
\[\Delta F_{N}\left(R,T_{s},x\right)\approx\frac{4\pi}{3}x^{3}n_{S}\left(-\Delta E +T_{s}\cdot\Delta S-\frac{\Delta n}{n_{S}}\frac{2\gamma_{L}}{3n_{L}R}\right)+4 \pi x^{2}\gamma_{SL} \tag{6}\]  
where \(\gamma_{SL}\) is the solid-liquid interface energy, \(\Delta n=n_{S}-n_{L}\). The change in entropy upon melting \(\Delta S\) as expressed using Equation (3) depends on the NP radius \(R\); namely, it decreases when decreasing \(R\). By substituting it into Equation (6), we obtain  
\[\Delta F_{N}\left(R,T,x\right)\approx\frac{4\pi}{3}x^{3}n_{S}\Delta E\left(- \delta\cdot\left(1-\frac{\alpha}{R}\right)-\frac{\alpha}{R}\left(1+\frac{ \gamma_{L}}{\Delta\gamma}\frac{\Delta n}{n_{S}}\right)\right)+4\pi x^{2}\gamma _{SL} \tag{7}\]  
where \(\delta=1-T_{s}/T_{m}\), \(\alpha=2\Delta\gamma/(n_{L}\Delta E)\), and \(\Delta n=n_{S}-n_{L}\). The parameter \(\delta\) denotes relative thermal hysteresis for the NP. The nucleation barrier is defined as \(\Delta F(R,T_{s},x=x_{\rm m})\), where \(x_{m}=2\gamma_{SL}\cdot\left[n_{S}\Delta E\left(\delta\cdot\left(1-\frac{ \alpha}{R}\right)+\frac{\alpha}{R}\left(1+\frac{\gamma_{L}}{\Delta\gamma}\frac {\Delta n}{n_{S}}\right)\right)\right]^{-1}\), and is equal to  
\[\Delta F_{N}(R,T_{s})\approx\frac{2\pi}{3}\frac{\left(2\gamma_{SL}\right)^{3}} {\left(\Delta E\right)^{2}n_{S}^{2}\left(\delta\cdot\left(1-\frac{\alpha}{R} \right)+\frac{\alpha}{R}\left(1+\frac{\gamma_{L}}{\Delta\gamma}\frac{\Delta n }{n_{S}}\right)\right)^{2}} \tag{8}\]  
It is worth pointing out that considering 0\(<\)\(\delta\)\(<\)1, the nucleation barrier given by Equation (8) decreases with a decreasing particle radius. To study the solidification of NPs as a function of their size, we use Equation (5) for the reference particle with the radius \(R_{0}\), which is large enough that the terms \(\sim 1/R_{0}\) in the corresponding nucleation barrier can be neglected. By substituting Equation (4) into Equation (5), for which \(\Delta F_{N}(R,T_{s})\) has been given by Equation (8), and omitting the terms that are identical for both actual and reference particles, we obtain  
\[-\frac{4\pi}{3}\frac{\alpha^{2}(1-\lambda)^{2}}{\beta(1-\delta)\cdot\left( \delta\cdot\left(1-\frac{\alpha}{R}\right)+\frac{\alpha}{R}(1+6.7\lambda) \right)^{2}}+3\ln(R)\approx-\frac{4\pi}{3}\frac{\alpha_{0}^{2}(1-\lambda)^{2} }{\beta_{0}(1-\delta_{0})\delta_{0}^{2}}+3\ln(R_{0}) \tag{9}\]  
where \(\beta=k_{B}T_{m}/\gamma_{SL}\), \(\beta_{0}=k_{B}T_{s0}/\gamma_{SL0}\), \(\alpha_{0}=2\Delta\gamma_{0}/(n_{L}\Delta E_{0})\), \(\lambda=1-n_{L}/n_{S}\), and the parameters with '0' in the subscript correspond to the reference particle with the melting temperature \(T_{m0}\) and solidification temperature \(T_{s0}\), i.e., for which relative thermal hysteresis \(\delta_{0}=1-T_{s0}/T_{m0}\). The logarithmic terms represent the R-dependent fractions of the nucleation sites' numbers. For illustrative purposes, in Equation (9), we also assumed that \(\Delta\gamma\approx\gamma_{SL}\approx 0.15\gamma_{L}\)[20].