2021
DOI: 10.1088/0256-307x/38/2/026501
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Scaling Behavior between Heat Capacity and Thermal Expansion in Solids

Abstract: We experimentally analyze the heat capacity and thermal expansion of reference solids in a wide temperature range from several Kelvin to melting temperature, and establish a universal double-linear relation between the experimental heat capacity C p and thermal expansion β, which is different from the previous models. The universal behavior between heat capacity and thermal expansion is important to predict the thermodynamic parameters at constant pressure, and is helpful for understanding th… Show more

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Cited by 8 publications
(10 citation statements)
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“…The parameters of K, V, and γ for Cu at room temperature are 140 GPa, 7.11 g/cm 3 , and 1.96, respectively. 18,28 The calculated value of [C 0 KV/(3γR) + E] for Cu is 499.6 kJ/mol, which is close to the fitted value 499.7 kJ/ mol reported in reference 18. Therefore, eq 4 is a universal equation to describe the experimental heat capacity C p of solids below melting temperature.…”
Section: ■ Model and Discussionsupporting
confidence: 85%
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“…The parameters of K, V, and γ for Cu at room temperature are 140 GPa, 7.11 g/cm 3 , and 1.96, respectively. 18,28 The calculated value of [C 0 KV/(3γR) + E] for Cu is 499.6 kJ/mol, which is close to the fitted value 499.7 kJ/ mol reported in reference 18. Therefore, eq 4 is a universal equation to describe the experimental heat capacity C p of solids below melting temperature.…”
Section: ■ Model and Discussionsupporting
confidence: 85%
“…18 The experimental heat capacity C p approximately shows different linear dependencies on the β above or below 3R in the reference solid materials (as seen in Figure 2), which can be seen in the previous study. 18 The non-volume-dependent heat capacity C T below T m is calculated by the Debye model (C 0 f D, green lines shown in Figure 4). The maximum of C T is the fitting parameter C 0 in eq 2, and the Debye temperature θ is determined at half of the parameter C 0 : f D (θ/T = 4.02) = 0.5.…”
Section: ■ Model and Discussionsupporting
confidence: 67%
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