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Difficulty: MediumThermal Expansion of Liquids and Anomalous Expansion of Water

A glass relative density bottle with linear expansivity α=8.0×10−6 K−1\alpha = 8.0 \times 10^{-6}\text{ K}^{-1} is completely filled with 204 g204\text{ g} of a liquid at 15∘C15^\circ\text{C}. When heated to 65∘C65^\circ\text{C}, 4 g4\text{ g} of the liquid overflows. What is the real cubic expansivity of the liquid?

  1. 4.24×10−4 K−14.24 \times 10^{-4}\text{ K}^{-1}Answer
  2. B
    4.08×10−4 K−14.08 \times 10^{-4}\text{ K}^{-1}
  3. C
    4.00×10−4 K−14.00 \times 10^{-4}\text{ K}^{-1}
  4. D
    3.76×10−4 K−13.76 \times 10^{-4}\text{ K}^{-1}

Answer

The real cubic expansivity of the liquid is 4.24×10−4 K−14.24 \times 10^{-4}\text{ K}^{-1}.
The real cubic expansivity of a liquid accounts for both the apparent expansion observed as overflow and the expansion of the container itself. Calculating apparent cubic expansivity yields γa=4 g200 g×50 K=4.0×10−4 K−1\gamma_a = \frac{4\text{ g}}{200\text{ g} \times 50\text{ K}} = 4.0 \times 10^{-4}\text{ K}^{-1}. Combining this with the glass bottle's cubic expansivity γv=3×8.0×10−6=0.24×10−4 K−1\gamma_v = 3 \times 8.0 \times 10^{-6} = 0.24 \times 10^{-4}\text{ K}^{-1} gives γr=4.24×10−4 K−1\gamma_r = 4.24 \times 10^{-4}\text{ K}^{-1}.

Step-by-Step Solution

1
Calculate the mass of liquid remaining in the bottle at the higher temperature.
mremaining=204 g−4 g=200 gm_{\text{remaining}} = 204\text{ g} - 4\text{ g} = 200\text{ g}
The apparent expansivity formula using the mass method requires the remaining mass of liquid that occupies the bottle volume at the elevated temperature.
2
Calculate the apparent cubic expansivity (γa\gamma_a) of the liquid.
γa=mass expelledmremaining×ΔT=4200×(65−15)=410000=4.0×10−4 K−1\gamma_a = \frac{\text{mass expelled}}{m_{\text{remaining}} \times \Delta T} = \frac{4}{200 \times (65 - 15)} = \frac{4}{10000} = 4.0 \times 10^{-4}\text{ K}^{-1}
Apparent expansivity is defined as the mass of liquid expelled divided by the product of remaining mass and temperature rise.
3
Determine the cubic expansivity of the glass vessel (γv\gamma_v).
γv=3α=3×(8.0×10−6 K−1)=2.4×10−5 K−1=0.24×10−4 K−1\gamma_v = 3\alpha = 3 \times (8.0 \times 10^{-6}\text{ K}^{-1}) = 2.4 \times 10^{-5}\text{ K}^{-1} = 0.24 \times 10^{-4}\text{ K}^{-1}
The cubic expansivity of an isotropic solid container is three times its linear expansivity.
4
Calculate the real cubic expansivity of the liquid (γr\gamma_r).
γr=γa+γv=4.0×10−4 K−1+0.24×10−4 K−1=4.24×10−4 K−1\gamma_r = \gamma_a + \gamma_v = 4.0 \times 10^{-4}\text{ K}^{-1} + 0.24 \times 10^{-4}\text{ K}^{-1} = 4.24 \times 10^{-4}\text{ K}^{-1}
The real volume expansivity of a liquid equals the sum of its apparent expansivity and the cubic expansivity of its container.

Key Concept

Relationship between real cubic expansivity, apparent cubic expansivity, and vessel cubic expansivity (γr=γa+γv\gamma_r = \gamma_a + \gamma_v).
Estimated Time:1m 30s
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Topics

Thermal Expansion of Liquids and Anomalous Expansion of Water

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