Examkin

Question

Difficulty: Very hardThermal Expansion of Liquids and Anomalous Expansion of Water

A glass vessel with a linear expansivity of 9.0×10−6 K−19.0 \times 10^{-6} \text{ K}^{-1} has a volume capacity of 300 cm3300 \text{ cm}^3 at 15∘C15^\circ\text{C}. It is completely filled with a liquid at this temperature. When the system is uniformly heated to 65∘C65^\circ\text{C}, a volume of 4.5 cm34.5 \text{ cm}^3 of the liquid overflows. What is the real cubic expansivity of the liquid?

  1. 3.27×10−4 K−13.27 \times 10^{-4} \text{ K}^{-1}Answer
  2. B
    3.09×10−4 K−13.09 \times 10^{-4} \text{ K}^{-1}
  3. C
    3.00×10−4 K−13.00 \times 10^{-4} \text{ K}^{-1}
  4. D
    2.73×10−4 K−12.73 \times 10^{-4} \text{ K}^{-1}

Answer

The real cubic expansivity of the liquid is 3.27×10−4 K−13.27 \times 10^{-4} \text{ K}^{-1}.
The real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubical expansivity of the containing vessel. The apparent cubic expansivity is 4.5300×50=3.0×10−4 K−1\frac{4.5}{300 \times 50} = 3.0 \times 10^{-4} \text{ K}^{-1}. The cubical expansivity of the vessel is 3×9.0×10−6=2.7×10−5 K−13 \times 9.0 \times 10^{-6} = 2.7 \times 10^{-5} \text{ K}^{-1}. Summing these yields 3.27×10−4 K−13.27 \times 10^{-4} \text{ K}^{-1}.

Step-by-Step Solution

1
Calculate the temperature change (ΔT\Delta T) and the apparent cubic expansivity (γa\gamma_a) of the liquid.
ΔT=65∘C−15∘C=50 K\Delta T = 65^\circ\text{C} - 15^\circ\text{C} = 50 \text{ K}. γa=VoverflowV0⋅ΔT=4.5300×50=3.0×10−4 K−1\gamma_a = \frac{V_{\text{overflow}}}{V_0 \cdot \Delta T} = \frac{4.5}{300 \times 50} = 3.0 \times 10^{-4} \text{ K}^{-1}.
Apparent expansion measures the overflow relative to initial volume and temperature rise.
2
Calculate the cubical expansivity of the glass vessel (γv\gamma_v).
γv=3αv=3×(9.0×10−6 K−1)=2.7×10−5 K−1=0.27×10−4 K−1\gamma_v = 3 \alpha_v = 3 \times (9.0 \times 10^{-6} \text{ K}^{-1}) = 2.7 \times 10^{-5} \text{ K}^{-1} = 0.27 \times 10^{-4} \text{ K}^{-1}.
The volume (cubical) expansivity of an isotropic solid container is three times its linear expansivity.
3
Determine the real cubic expansivity of the liquid (γr\gamma_r) using the relation γr=γa+γv\gamma_r = \gamma_a + \gamma_v.
γr=3.0×10−4 K−1+0.27×10−4 K−1=3.27×10−4 K−1\gamma_r = 3.0 \times 10^{-4} \text{ K}^{-1} + 0.27 \times 10^{-4} \text{ K}^{-1} = 3.27 \times 10^{-4} \text{ K}^{-1}.
Real volume expansion equals the observed apparent expansion plus the expansion of the container.

Key Concept

Relationship between real expansivity, apparent expansivity, and vessel expansivity: γr=γa+γv\gamma_r = \gamma_a + \gamma_v, where γv=3α\gamma_v = 3\alpha.
Rate this question

Topics

Thermal Expansion of Liquids and Anomalous Expansion of Water

More questions from this topic