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Difficulty: HardThermal Expansion of Liquids and Anomalous Expansion of Water

A metallic container with an initial volume of 400 cm3400 \text{ cm}^3 at 15∘C15^\circ\text{C} is filled completely with paraffin. Upon heating the container and its contents to 65∘C65^\circ\text{C}, a volume of 18 cm318 \text{ cm}^3 of paraffin spills over. Given that the linear expansivity of the metal container is 2.0×10−5 K−12.0 \times 10^{-5} \text{ K}^{-1}, what is the real cubic expansivity of the paraffin, expressed in units of 10−4 K−110^{-4} \text{ K}^{-1}?

Answer: 9.6 10^-4 K^-1

Answer

The real cubic expansivity of the paraffin is 9.6×10−4 K−19.6 \times 10^{-4} \text{ K}^{-1}, which corresponds to a numerical value of 9.69.6 in units of 10−4 K−110^{-4} \text{ K}^{-1}.
The real cubic expansivity of a liquid accounts for both the observed (apparent) expansion of the liquid and the expansion of the container holding it. By applying γa=ΔVaV0ΔT=9.0×10−4 K−1\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = 9.0 \times 10^{-4} \text{ K}^{-1} and γv=3α=0.6×10−4 K−1\gamma_v = 3\alpha = 0.6 \times 10^{-4} \text{ K}^{-1}, we sum them to obtain the real cubic expansivity γr=9.6×10−4 K−1\gamma_r = 9.6 \times 10^{-4} \text{ K}^{-1}.

Step-by-Step Solution

1
Determine the temperature change (ΔT\Delta T) and the apparent change in volume (ΔVa\Delta V_a).
ΔT=65∘C−15∘C=50 K\Delta T = 65^\circ\text{C} - 15^\circ\text{C} = 50 \text{ K} and ΔVa=18 cm3\Delta V_a = 18 \text{ cm}^3.
The overflow volume represents the apparent expansion of the liquid relative to the expanding container over the temperature rise.
2
Calculate the apparent cubic expansivity (γa\gamma_a) of the paraffin.
γa=ΔVaV0ΔT=18400×50=1820000=9.0×10−4 K−1\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = \frac{18}{400 \times 50} = \frac{18}{20000} = 9.0 \times 10^{-4} \text{ K}^{-1}.
Apparent expansivity relates the apparent volume expansion to the original volume and temperature increase.
3
Calculate the cubic expansivity of the metallic vessel (γv\gamma_v).
γv=3α=3×2.0×10−5 K−1=6.0×10−5 K−1=0.6×10−4 K−1\gamma_v = 3 \alpha = 3 \times 2.0 \times 10^{-5} \text{ K}^{-1} = 6.0 \times 10^{-5} \text{ K}^{-1} = 0.6 \times 10^{-4} \text{ K}^{-1}.
Cubic expansivity of an isotropic solid container is three times its linear expansivity.
4
Calculate the real cubic expansivity of the paraffin (γr\gamma_r).
γr=γa+γv=9.0×10−4 K−1+0.6×10−4 K−1=9.6×10−4 K−1\gamma_r = \gamma_a + \gamma_v = 9.0 \times 10^{-4} \text{ K}^{-1} + 0.6 \times 10^{-4} \text{ K}^{-1} = 9.6 \times 10^{-4} \text{ K}^{-1}.
The real expansion of a liquid is the sum of its apparent expansion and the expansion of the containing vessel.

Key Concept

Real vs. Apparent Expansion of Liquids (γr=γa+γv\gamma_r = \gamma_a + \gamma_v where γv=3α\gamma_v = 3\alpha)
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Topics

Thermal Expansion of Liquids and Anomalous Expansion of Water

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