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Chapter 18: Problem 19

A vertical glass tube of length \(L=1.280000 \mathrm{~m}\) is half filled with a liquid at \(20.000000^{\circ} \mathrm{C}\). How much will the height of the liquid column change when the tube and liquid are heated to \(30.000000^{\circ} \mathrm{C} ?\) Use coefficients \(\alpha_{\text {glass }}=1.000000 \times 10^{-5} / \mathrm{K}\) and \(\beta_{\text {liquid }}=4.000000 \times 10^{-5} / \mathrm{K}\) .

Short Answer

Expert verified
The height of the liquid column will increase by 0.000128 m.

Step by step solution

01

Determine the initial height of the liquid

The initial height of the liquid column is half the length of the tube. Therefore, the initial height is \( h_0 = \frac{L}{2} = \frac{1.280000}{2} = 0.640000 \text{ m} \).
02

Calculate the change in height of the glass tube

The change in height of the glass tube is determined by its linear expansion. The formula for linear expansion is \( \Delta L = \alpha \cdot L_0 \cdot \Delta T \), where \( \Delta T \) is the temperature change.
03

Compute the temperature change

The temperature change, \( \Delta T \), is calculated by subtracting the initial temperature from the final temperature. Thus, \( \Delta T = 30.000000^{\circ} \mathrm{C} - 20.000000^{\circ} \mathrm{C} = 10.000000 \mathrm{K} \).
04

Calculate the change in length of the tube

Using the formula from Step 2, with \( \alpha = 1.000000 \times 10^{-5} / \mathrm{K} \), we find:\[\Delta L_{\text{glass}} = (1.000000 \times 10^{-5}) \times 1.280000 \times 10.000000 = 0.000128 \text{ m}.\]
05

Determine the change in height of the glass tube's liquid capacity

Since the glass tube itself expands, the new effective height for the liquid is \( h_{\text{new, glass}} = h_0 + \Delta L_{\text{glass}} = 0.640000 + 0.000128 = 0.640128 \text{ m}. \)
06

Calculate the change in height of the liquid column

The volume expansion of the liquid can be calculated using \( \Delta V = \beta \cdot V_0 \cdot \Delta T \). Since the original volume is \( V_0 = A \cdot h_0 \), where \( A \) is the cross-sectional area, the change in height of the liquid is \( \Delta h = \Delta V / A \).
07

Compute the change in height of the liquid

For volumetric expansion, use \( \beta = 4.000000 \times 10^{-5} / \mathrm{K} \):\[\Delta h_{\text{liquid}} = (4.000000 \times 10^{-5}) \times 0.640000 \times 10.000000 = 0.000256 \text{ m}.\]
08

Determine the total change in height of the liquid column

The total height change of the liquid column accounts for the expansion changes of both the tube and the liquid:\[\Delta h_{\text{total}} = \Delta h_{\text{liquid}} - \Delta L_{\text{glass}} = 0.000256 - 0.000128 = 0.000128 \text{ m}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Expansion
When materials such as solids are subjected to changes in temperature, they expand or contract. This is known as thermal expansion. Linear expansion specifically refers to the change in length of a material as its temperature changes. The degree of expansion is influenced by the material's thermal coefficient of linear expansion, denoted as \( \alpha \). The linear expansion can be calculated using the formula:
\[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \]where:
  • \( \Delta L \) is the change in length.
  • \( \alpha \) is the coefficient of linear expansion.
  • \( L_0 \) is the original length.
  • \( \Delta T \) is the temperature change.
In our exercise, the glass tube experiences linear expansion as it is heated. This results in an increase in its length, altering the apparent volume it can hold.
Volumetric Expansion
Volumetric expansion refers to the change in volume of a substance when its temperature changes. It is more pertinent to liquids and gases, as they don't just expand in one dimension (length) but expand in all three dimensions, increasing their volume. The formula to find the volumetric expansion is:
\[ \Delta V = \beta \cdot V_0 \cdot \Delta T \]where:
  • \( \Delta V \) is the change in volume.
  • \( \beta \) is the coefficient of volumetric expansion.
  • \( V_0 \) is the original volume.
  • \( \Delta T \) is the temperature change.
In this context, as the liquid inside the tube gets warmer, it expands volumetrically. This expansion leads to an increase in the height of the liquid column, given that the container's other dimensions remain constrained by it being a glass tube.
Thermal Coefficient
The thermal coefficient is a crucial factor in determining how much a material will expand or contract with temperature shifts. There are two types of coefficients frequently used:
  • Coefficient of Linear Expansion (\( \alpha \)): This is used for solid materials and influences how much the length changes with temperature.
  • Coefficient of Volumetric Expansion (\( \beta \)): This is applied to liquids and gases, affecting their volume change.
The value of a thermal coefficient helps predict how substances behave under temperature variations.
For example, in the exercise, different thermal coefficients are provided for the glass and the liquid, guiding us to calculate their expansion. A higher thermal coefficient implies a more significant expansion for the material when subjected to the same temperature change.

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Most popular questions from this chapter

The area \(A\) of a rectangular plate is \(a b=1.4 \mathrm{~m}^{2} .\) Its coefficient of linear expansion is \(\alpha=32 \times 10^{-6} / \mathrm{C}^{\circ} .\) After a temperature rise \(\Delta T=89^{\circ} \mathrm{C}\), side \(a\) is longer by \(\Delta a\) and side \(b\) is longer by \(\Delta b\) (Fig. 18-61). Neglecting the small quantity \((\Delta a \Delta b) / a b\), find \(\Delta A\).

An insulated Thermos contains \(130 \mathrm{~cm}^{3}\) of hot coffee at \(80.0^{\circ} \mathrm{C}\). You put in a \(12.0 \mathrm{~g}\) ice cube at its melting point to cool the coffee. By how many degrees has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment.

The specific heat of a substance varies with temperature according to the function \(c=0.20+0.14 T+0.023 T^{2}\), with \(T\) in \({ }^{\circ} \mathrm{C}\) and \(c\) in cal \(/ \mathrm{g} \cdot \mathrm{K}\). Find the energy required to raise the temperature of \(2.0 \mathrm{~g}\) of this substance from \(5.0^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\).

In a solar water heater, energy from the Sun is gathered by water that circulates through tubes in a rooftop collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes; this water is pumped into a holding tank. Assume that the efficiency of the overall system is \(20 \%\) (that is, \(80 \%\) of the incident solar energy is lost from the system). What collector area is necessary to raise the temperature of 200 \(\mathrm{L}\) of water in the tank from \(20^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) in \(1.0 \mathrm{~h}\) when the intensity of incident sunlight is \(700 \mathrm{~W} / \mathrm{m}^{2}\) ?

What mass of butter, which has a usable energy content of \(6.0 \mathrm{Cal} / \mathrm{g}(=6000 \mathrm{cal} / \mathrm{g})\), would be equivalent to the change in gravitational potential energy of a \(73.0 \mathrm{~kg}\) man who ascends from sea level to the top of Mt. Everest, at elevation \(8.84 \mathrm{~km}\) ? Assume that the average \(g\) for the ascent is \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).
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