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Zorluk: KolayThermal Expansion of Liquids and Anomalous Expansion of Water

A liquid has a real cubic expansivity of 5.0×104 K15.0 \times 10^{-4} \text{ K}^{-1}. When this liquid is heated inside a metallic vessel, its apparent cubic expansivity is determined to be 4.1×104 K14.1 \times 10^{-4} \text{ K}^{-1}. What is the linear expansivity of the metallic vessel in 105 K110^{-5} \text{ K}^{-1}?

Cevap: 3 10^-5 K^-1

Cevap

The linear expansivity of the metallic vessel is 3.0×105 K13.0 \times 10^{-5} \text{ K}^{-1}.
Real cubic expansivity of a liquid accounts for both the expansion of the liquid relative to the container and the expansion of the container itself: γr=γa+γv\gamma_r = \gamma_a + \gamma_v. Subtracting the apparent cubic expansivity (4.1×104 K14.1 \times 10^{-4} \text{ K}^{-1}) from the real cubic expansivity (5.0×104 K15.0 \times 10^{-4} \text{ K}^{-1}) gives the vessel's cubic expansivity of 0.9×104 K1=9.0×105 K10.9 \times 10^{-4} \text{ K}^{-1} = 9.0 \times 10^{-5} \text{ K}^{-1}. Dividing this value by 3 gives the vessel's linear expansivity: α=3.0×105 K1\alpha = 3.0 \times 10^{-5} \text{ K}^{-1}.

Adım Adım Çözüm

1
Find the cubic expansivity of the vessel (γv\gamma_v)
γv=γrγa=5.0×1044.1×104=0.9×104 K1=9.0×105 K1\gamma_v = \gamma_r - \gamma_a = 5.0 \times 10^{-4} - 4.1 \times 10^{-4} = 0.9 \times 10^{-4} \text{ K}^{-1} = 9.0 \times 10^{-5} \text{ K}^{-1}
The real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubic expansivity of the containing vessel (γr=γa+γv\gamma_r = \gamma_a + \gamma_v).
2
Calculate the linear expansivity of the vessel (α\alpha)
α=γv3=9.0×1053=3.0×105 K1\alpha = \frac{\gamma_v}{3} = \frac{9.0 \times 10^{-5}}{3} = 3.0 \times 10^{-5} \text{ K}^{-1}
For isotropic solids, volume (cubic) expansivity is three times the linear expansivity (γv=3α\gamma_v = 3\alpha).

Anahtar Kavram

Relationship between real cubic expansivity, apparent cubic expansivity, and vessel expansivity
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