On a phase diagram, the eutectic temperature is seen as the eutectic point (see plot).
Pada diagram fase di samping, persimpangan suhu eutektik dan komposisi eutektik memberikan titik eutektik.
There is also a thermal arrest for the duration of the phase change during which the temperature of the system does not change.
Selama fase perubahan ini, terdapat juga penangkapan termal di mana suhu dari sistem tidak berubah.
The composition and temperature of a eutectic can be calculated from enthalpy and entropy of fusion of each components.
Komposisi dan suhu eutektik dapat dihitung dari entalpi dan entropi fusi masing-masing komponen.
=\mu _{i}^{\circ }+RT\ln {\frac {a_{i}}{a}}\approx \mu _{i}^{\circ }+RT\ln x_{i}.}
=\mu _{i}^{\circ }+RT\ln x_{i}=0\Rightarrow \mu _{i}^{\circ }=-RT\ln x_{i}.}
=\mu _{i}^{\circ }+RT\ln x_{i}=0\Rightarrow \mu _{i}^{\circ }=-RT\ln x_{i}.} Using and integrating gives ( ∂ μ i / T ∂ T ) P∂ ∂ T ( R ln ⁡ x i ) ⇒ R ln ⁡ x i− H i ∘ T + K .
μ iμ i ∘ + R T ln ⁡ a i a ≈ μ i ∘ + R T ln ⁡ x i\mu _{i}^{\circ }+RT\ln {\frac {a_{i}}{a}}\approx \mu _{i}^{\circ }+RT\ln x_{i}} Pada kesetimbangan, μ i00} , thus μ i ∘ ^{\circ }} didapat dengan: μ iμ i ∘ + R T ln ⁡ x i0 ⇒ μ i ∘− R T ln ⁡ x i .
/T}{\partial T}}\right)_{P}={\frac {\partial }{\partial T}}\left(R\ln x_{i}\right)\Rightarrow R\ln x_{i}=-{\frac {H_{i}^{\circ }}{T}}+K.} The integration constant K may be determined for a pure component with a melting temperature T ∘ } and an enthalpy of fusion H ∘ } :
x i1 ⇒ TT i ∘ ⇒ KH i ∘ T i ∘1\Rightarrow T=T_{i}^{\circ }\Rightarrow K={\frac {H_{i}^{\circ }}{T_{i}^{\circ }}}} Kita mendapatkan hubungan yang menentukan fraksi molar sebagai fungsi suhu untuk masing-masing komponen.
x i1 ⇒ TT i ∘ ⇒ KH i ∘ T i ∘ .1\Rightarrow T=T_{i}^{\circ }\Rightarrow K={\frac {H_{i}^{\circ }}{T_{i}^{\circ }}}.} We obtain a relation that determines the molar fraction as a function of the temperature for each component: R ln ⁡ x i− H i ∘ T + H i ∘ T i ∘ .
x i1 ⇒ TT i ∘ ⇒ KH i ∘ T i ∘1\Rightarrow T=T_{i}^{\circ }\Rightarrow K={\frac {H_{i}^{\circ }}{T_{i}^{\circ }}}} Kita mendapatkan hubungan yang menentukan fraksi molar sebagai fungsi suhu untuk masing-masing komponen.
i < n ⇒ ln ⁡ x i + H i ∘ R T − H i ∘ R T i ∘0 , ln ⁡ ( 1 − ∑ i1 n − 1 x i ) + H n ∘ R T − H n ∘ R T n ∘0 , \forall i<n\Rightarrow \ln x_{i}+{\frac {H_{i}^{\circ }}{RT}}-{\frac {H_{i}^{\circ }}{RT_{i}^{\circ }}}=0,\\\ln \left(1-\sum \limits _{i=1}^{n-1}x_{i}\right)+{\frac {H_{n}^{\circ }}{RT}}-{\frac {H_{n}^{\circ }}{RT_{n}^{\circ }}}=0,\end{cases}}} which can be solved by Azeotrope, or constant boiling mixture Freezing-point depression Fusible alloy Smith, William F.; Hashemi, Javad (2006), Foundations of Materials Science and Engineering (4th ed.), McGraw-Hill, ISBN 978-0-07-295358-9.
i ∘0 ln ⁡ ( 1 − ∑ i1 n − 1 x i ) + H n ∘ R T − H n ∘ R T n ∘0 {l}\left\{{\begin{array}{*{20}c}{\forall i<n\Rightarrow \ln x_{i}+{\frac {H_{i}^{\circ }}{RT}}-{\frac {H_{i}^{\circ }}{RT_{i}^{\circ }}}=0}\\{\ln \left({1-\sum \limits _{i=1}^{n-1}{x_{i}}}\right)+{\frac {H_{n}^{\circ }}{RT}}-{\frac {H_{n}^{\circ }}{RT_{n}^{\circ }}}=0}\\\end{array}}\right.\\\\\end{array}}} yang dapat diselesaikan dengan
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