Theoretical Analysis of Melting Point Depression of Pure Metals in Different Initial Configurations

Abstract

A general equation is derived for melting point depression (MPD) of pure metals, consisting of three terms: MPD due to high gas pressure, MPD due to high strain energy, and MPD due to small size of the metal. Particular equations are derived for different configurations of the solid metal, including grains embedded within a matrix. The equations obtained in this paper can be used to design nano-joining structures with improved MPD.

Appendices

Appendix 1: Physical Properties for the Cu/AlN System

The standard melting point of Cu is T ^o_m  = 1358.0 ± 0.2 K (Ref 53). The standard molar melting entropy is Δ_m S ^o_mol  = 9.7 ± 0.3 J/mol K (Ref 53). The molar volume (m³/mol) of solid copper is written by the equation (Ref 54, 55)

$$ V_{\text{s}}^{\text{o}} = 7.011 \times 10^{ - 6} + 2.518 \times 10^{ - 10} T + 1.178 \times 10^{ - 13} T^{2} \pm 1\% . $$

(A1)

The molar volume of solid Cu at 298 K and at its standard melting point is (7.10 ± 0.07) 10⁻⁶ and (7.57 ± 0.08) 10⁻⁶ m³/mol, respectively. The bulk packing fraction of the fcc crystal is 0.74. From here, the standard atomic radius of Cu is r _a = 0.128 nm. The linear heat expansion coefficient (1/K) of the solid Cu follows from Eq A1 as

$$ \upalpha_{\text{s}} \equiv \frac{1}{{3V_{\text{s}}^{\text{o}} }}\frac{{dV_{\text{s}}^{\text{o}} }}{dT} = \frac{{8.39 \times 10^{ - 11} + 7.85 \times 10^{ - 14} T}}{{7.011 \times 10^{ - 6} + 2.518 \times 10^{ - 10} \,T + 1.178 \times 10^{ - 13} \,T^{2} }} \pm 2\% . $$

(A2)

The linear heat expansion coefficient of solid Cu at 298 K and at its standard melting point is (1.51 ± 0.03) 10⁻⁵ and (2.52 ± 0.05) 10⁻⁵ 1/K, respectively, the average value is (2.0 ± 0.5) 10⁻⁵ 1/K. The molar volume (m³/mol) of liquid copper is written by the equation (Ref 54-56)

$$ V_{\text{l}}^{\text{o}} = 7. 10 2 \times 10^{ - 6} + 4. 3 7 7 \times 10^{ - 10} T + 1. 3 2 5 \times 10^{ - 1 3} T^{ 2} \pm 2\% . $$

(A3)

The molar volume of liquid Cu at 298 K and at its standard melting point is (7.24 ± 0.14) 10⁻⁶ and (7.94 ± 0.16) 10⁻⁶ m³/mol, respectively. The linear heat expansion coefficient (1/K) of liquid Cu follows from Eq A3 as

$$ \upalpha_{l} \equiv \frac{1}{{3V_{\text{l}}^{\text{o}} }}\frac{{dV_{\text{l}}^{\text{o}} }}{dT} = \frac{{1.46 \times 10^{ - 10} + 8.83 \times 10^{ - 14} T}}{{7.102 \times 10^{ - 6} + 4.377 \times 10^{ - 10} T + 1.325 \times 10^{ - 13} T^{2} }} \pm 2\% . $$

(A4)

The linear heat expansion coefficient of liquid Cu at 298 K and at its standard melting point is (2.38 ± 0.05) 10⁻⁵ and (3.35 ± 0.07) 10⁻⁵ 1/K, respectively, the average value is (2.9 ± 0.5) 10⁻⁵ 1/K. The average molar volume (m³/mol) of the solid and liquid phases is

$$ V^{\text{o}} = 7.0565 \times 10^{ - 6} + 3.4475 \times 10^{ - 10} T + 1.2515 \times 10^{ - 13} T^{2} \pm 1\% . $$

(A5)

The molar volume change upon melting (m³/mol) is obtained as the difference between Eq A3 and A1, while the relative molar volume change upon melting is the latter divided by the molar volume of the solid

$$ \Delta_{\text{m}} V^{\text{o}} \cong 9.1 \times 10^{ - 8} + 1.859 \times 10^{ - 10} T + 1.47 \times 10^{ - 14} T^{2} , $$

(A6)

$$ \frac{{\Delta_{\text{m}} V^{\text{o}} }}{{V_{\text{s}}^{\text{o}} }} \cong \frac{{9.1 \times 10^{ - 8} + 1.859 \times 10^{ - 10} T + 1.47 \times 10^{ - 14} T^{2} }}{{7.011 \times 10^{ - 6} + 2.518 \times 10^{ - 10} T + 1.178 \times 10^{ - 13} T^{2} }}. $$

(A7)

The molar volume change upon melting at 298 K and at its standard melting point approximately equals 1.48 × 10⁻⁷ m³/mol (2.08%) and 3.71 × 10⁻⁷ m³/mol (4.90%), respectively.

The standard entropy of melting of Cu per unit volume at standard melting point of 1358 K is the ratio of Δ_m S ^o_mol  = 9.7 ± 0.3 J/mol K and the average molar volume of V ^o = (7.76 ± 0.12) 10⁻⁶ m³/mol: Δ_m S ^o = (1.25 ± 0.06) 10⁶ J/m³K.

The standard surface tension and surface energy of copper (Ref 56, 57) (J/m²)

$$ \upsigma_{{{\text{l}}/{\text{g}}}}^{\text{o}} = 1.30 - 2.3 \times 10^{ - 4} (T - T_{\text{m}}^{\text{o}} ), $$

(A8)

$$ \upsigma_{{{\text{s}}/{\text{g}}}}^{\text{o}} = 1.60 - 4.7 \times 10^{ - 4} (T - T_{\text{m}}^{\text{o}} ). $$

(A9)

The bulk modulus of liquid copper at its melting point is the inverse of its bulk compressibility and it is estimated as K _l = 100 GPa (Ref 56). It probably decreases linearly with the decrease of the cohesion energy in the liquid. The standard cohesion energy of liquid Cu at its melting point is about (Ref 57)

$$ H_{\text{Cu,l,m}}^{\text{o}} = - 26.3RT_{\text{m,Cu}}^{\text{o}} + 2.62 \times 10^{ - 4} (RT_{\text{m,Cu}}^{\text{o}} )^{2} , $$

(A10)

where R = 8.3145 J/mol K, the universal gas constant. Substituting T ^o_m  = 1358.0 K into Eq A10: H ^o_Cu,l,m  = −263 kJ/mol. The heat capacity of liquid Cu is about 32.8 J/mol K, and is not T-dependent (Ref 58). Thus, the T-coefficient of the cohesion energy change is the ratio of these two values: 1.25 × 10⁻⁴ 1/K. The final T-dependence of the bulk modulus of liquid Cu is

$$ K_{\text{l}} \cong 120 (1 - 1.25 \times 10^{ - 4} T)\,{\text{GPa}} . $$

(A11)

The T-dependence of solid copper is approximately written as (Ref 59)

$$ K_{\text{s}} \cong 143\,(1 - 1.4 \times 10^{ - 4} T)\,{\text{GPa}} . $$

(A12)

The Young modulus and Poisson ratio of AlN at 298 K are E _m = 308 GPa and ν_m = 0.25 (Ref 60). Thus, the shear modulus from Eq 31: μ_m = 120 GPa at 298 K. Its T-dependence scales with the melting point, so in comparison to solid Cu of Eq A12 its T-coefficient is lower by about a coefficient of 2.4

$$ \upmu_{\text{m}} \cong 122\,(1 - 5.8 \times 10^{ - 5} T)\,{\text{GPa,}} $$

(A13a)

$$ E_{\text{m}} \cong 313 (1 - 5.8 \times 10^{ - 5} T)\,{\text{GPa}} . $$

(A13b)

The average linear expansion coefficient of AlN is α_m = 5.27 × 10⁻⁶ 1/K (Ref 60) between 20 and 800°C. No temperature dependence is taken into account.

Calculations will be performed for T _o = 298 K. So, the two integrals of Eq 32 are described approximately as

$$ \int\limits_{{T_{\text{o}} }}^{{T_{\text{m}} }} {(\upalpha_{\text{s}} - \upalpha_{\text{m}} )dT} = - 2.56 \times 10^{ - 3} + 7.22 \times 10^{ - 6} T + 4.56 \times 10^{ - 9} T^{2} , $$

(A14a)

$$ \int\limits_{{T_{\text{o}} }}^{{T_{\text{m}} }} {(\upalpha_{\text{l}} - \upalpha_{\text{m}} )dT} = - 5.21 \times 10^{ - 3} + 1.61 \times 10^{ - 5} T + 4.56 \times 10^{ - 9} T^{2} . $$

(A14b)

The liquid/matrix interfacial energy is defined as (Ref 61)

$$ \upsigma_{{{\text{l}}/{\text{m}}}}^{\text{o}} = \upsigma_{{{\text{l}}/{\text{g}}}}^{\text{o}} + \upsigma_{{{\text{m}}/{\text{g}}}}^{\text{o}} - W_{{{\text{l}}/{\text{m}}}}^{\text{o}} , $$

(A15)

where W ^o_l/m (J/m²) is the adhesion energy at the liquid Cu/AlN matrix interface. At the melting point of Cu: σ ^o_l/g  = 1.30 J/m² (see Eq A8). Surface energy of AlN is about σ ^o_m/g  = 2.53 J/m² at 1273 K (Ref 62). The contact angle of liquid Cu on AlN, as covalent and non-reactive ceramic is about 150° (Ref 61). Thus, from the Young-Dupré equation the adhesion energy at the melting point of Cu is about W ^o_l/m  = 0.17 J/m². In the first approximation, this value is T-independent. Thus, at the melting point of Cu: σ ^o_l/m  = 3.66 J/m². Its T-dependence is approximately

$$ \upsigma_{\text{l/m}}^{\text{o}} = 3.66 - 5 \times 10^{ - 4} (T - T_{\text{m}}^{\text{o}} )\,{\text{J}}/{\text{m}}^{2} . $$

(A16)

Solid Cu is not coherent with AlN. Thus, its adhesion energy is similar to that of the liquid Cu: W ^o_s/m  = 0.17 J/m². Then, the solid/matrix interfacial energy approximately equals

$$ \upsigma_{{{\text{s}}/{\text{m}}}}^{\text{o}} = 3.96 - 8 \times 10^{ - 4} (T - T_{\text{m}}^{\text{o}} )\,{\text{J}}/{\text{m}}^{2} . $$

(A17)

Appendix 2: The Derivation of a General Equation for the Strain Energy Density

When an elastic body is deformed, work W is done (J). This work, stored in the body is the strain energy. When this strain energy is divided by the volume V of the body (m³), the strain energy density E is obtained (J/m³ = N/m² = Pa).

In the simplest case, let us apply a tensile force F (N) to a cylindrical body of initial length L (m) and of cross-sectional area A (m²). As a result, the body will be elongated (deformed) by x (m), what is called the absolute strain. The relative strain (called simply strain) is the ratio of the absolute elongation and the initial length: ɛ ≡ x/L (ɛ is dimensionless). As a result, the work done and stored within the body as strain energy is W = 0.5Fx (J). The stress in the body (σ, Pa) is the tensile force per unit cross-sectional area: σ ≡ F/A. Now, let us express x from ɛ ≡ x/L and F from σ ≡ F/A and substitute into the equation of the strain energy: W = 0.5Fx = 0.5σAɛL. Remember, that AL = V (the volume of the body) and E ≡ W/V. Then, from the above

$$ E = \frac{\upsigma \upvarepsilon }{2}. $$

(B1)

When tensile changes are small, strain is proportional to stress: ɛ ≅ σ/E _Y, where E _Y (Pa) is the modulus of elasticity, or the Young modulus. Expressing from here σ as function of ɛ (or vice versa) we can obtain two versions for the equations of the strain energy density

$$ E = \frac{{E_{\text{Y}} \upvarepsilon^{2} }}{2}, $$

(B2)

$$ E = \frac{{\upsigma^{2} }}{{2E_{\text{Y}} }}. $$

(B3)