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At \(20^{\circ} \mathrm{C}\), a rod is exactly \(20.05 \mathrm{~cm}\) long on a steel ruler. Both the rod and the ruler are placed in an oven at \(270^{\circ} \mathrm{C}\), where the rod now measures \(20.11 \mathrm{~cm}\) on the same ruler. What is the coefficient of thermal expansion for the material of which the rod is made?

Short Answer

Expert verified
The coefficient of thermal expansion for the rod is \(2.40 × 10^{-5}/^{\text{C}} \).

Step by step solution

01

- Define thermal expansion

Thermal expansion can be calculated using the formula \[ \text{thermal expansion} = \text{initial length} \times \text{coefficient of thermal expansion} \times \text{change in temperature} \]
02

- Identify known values

Identify the initial length \[ L_0 = 20.05 \text{ cm} \] the observed length at the higher temperature \[ L_1 = 20.11 \text{ cm} \] and the change in temperature \[ \text{Δ}T = 270^{ewline\text{C}} - 20^{ewline\text{C}} = 250^{ewline\text{C}} \]
03

- Determine the apparent expansion

Since the rod and ruler both expand, first calculate the apparent length increase measured by the ruler. \[ \text{Apparent Expansion} = L_1 - L_0 = 20.11\text{ cm} - 20.05\text{ cm} = 0.06\text{ cm} \]
04

- Consider the expansion of the ruler

Let the coefficient of thermal expansion for steel be \(α_{\text{steel}} = 12 \times 10^{-6}/^{\text{C}} \). The steel ruler expands by \[ (\text{change in length})_{\text{ruler}} = L_{\text{ruler}} \times α_{\text{steel}} \times ΔT \] where \[ (\text{change in length})_{\text{ruler}} = 20.05 \text{ cm} \times 12 \times 10^{-6}/^{\text{C}} \times 250^{\text{C}} = 0.06015 \text{ cm} \]
05

- Calculate actual expansion of the rod

The actual length increase of the rod is the apparent increase plus the ruler's increase: \[ 0.06 \text{ cm} + 0.06015 \text{ cm} = 0.12015 \text{ cm} \]
06

- Apply thermal expansion formula

Using the formula for linear thermal expansion, \[ ΔL = L_0 × α_{\text{rod}} × ΔT \] solve for \( α_{\text{rod}} \): \[ 0.12015 \text{ cm} = 20.05 \text{ cm} × α_{\text{rod}} × 250^{\text{C}} \] \[ α_{\text{rod}} = \frac{0.12015 \text{ cm}}{20.05 \text{ cm} × 250^{\text{C}}} = 2.40 × 10^{-5}/^{\text{C}} \]

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Key Concepts

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

coefficient of thermal expansion
The coefficient of thermal expansion (CTE) is a material property that measures how much a material expands when its temperature changes. It is typically denoted by the Greek letter alpha (α) and has units of inverse temperature \((1/^{\text{C}})\). CTE is a crucial factor in engineering and construction because it predicts how materials will behave when subjected to temperature variations.
To calculate the CTE for a given material, you can use the formula:
\[ \Delta L = L_0 \times \alpha \times \Delta T \]
where \( \Delta L \) is the change in length, \( L_0 \) is the original length, \( \alpha \) is the coefficient of thermal expansion, and \( \Delta T \) is the change in temperature.
Understanding the CTE helps in designing materials and structures that can withstand thermal stress without deforming or breaking.
linear expansion
Linear expansion describes the change in length of a material when heated or cooled. It is a specific type of thermal expansion that focuses solely on the change in one dimension (length).
Materials expand linearly when subjected to a uniform temperature increase. The formula for linear expansion is:
\[ \Delta L = L_0 \times \alpha \times \Delta T \]
This formula highlights that the change in length (\(\Delta L\)) is directly proportional to the original length (\(L_0\)), the coefficient of thermal expansion (\(\alpha\)), and the change in temperature (\(\Delta T\)).
Examples include metal rods, which lengthen when heated, and concrete slabs, which expand on hot days.
temperature change
Temperature change (\(\Delta T\)) is the difference between the initial and final temperatures of a material. It directly influences the amount of thermal expansion a material undergoes.
In the exercise, the temperature change is calculated as follows:
\( \Delta T = 270^{\text{C}} - 20^{\text{C}} = 250^{\text{C}} \).
When you know the temperature change, you can use it to determine the change in length of a material using its coefficient of thermal expansion. A larger temperature change results in more significant expansion or contraction.
length measurement
Accurately measuring the length of materials before and after temperature changes is essential for calculating thermal expansion. In the given exercise, measurements were taken with a steel ruler.
Initially, the rod was measured to be
20.05 cm at 20°C. After heating to 270°C, the rod measured 20.11 cm.
These measurements allow you to determine the apparent expansion:
\( 20.11 \text{ cm} - 20.05 \text{ cm} = 0.06 \text{ cm} \).
However, because the steel ruler also expands, you must account for the ruler's expansion to find the accurate length change of the rod. This involves calculating the change in length of the ruler at the higher temperature.
material properties
Different materials have unique properties that determine their behavior under temperature changes. One key property is the coefficient of thermal expansion (CTE).
In the exercise, a steel ruler with a known CTE of \(12 \times 10^{-6}/^{\text{C}}\) was used. Steel’s expansion affects the accuracy of length measurements.
Material properties like thermal conductivity, specific heat, and elasticity also play roles in thermal expansion. Engineers must consider these properties when selecting materials for applications involving temperature fluctuations.
In summary, understanding material properties is vital for predicting and controlling thermal expansion in various engineering contexts.

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Most popular questions from this chapter

A small electric immersion heater is used to heat \(100 \mathrm{~g}\) of water for a cup of instant coffee. The heater is labeled "200 watts," so it converts electrical energy to thermal energy that is transferred to the water at this rate. Calculate the time required to bring the water from \(23^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) ignoring any thermal energy that transfers out of the cup.

A person makes a quantity of iced tea by mixing 500 \(\mathrm{g}\) of hot tea (essentially water) with an equal mass of ice at its melting point. If the initial hot tea is at a temperature of (a) \(90^{\circ} \mathrm{C}\) and (b) \(70^{\circ} \mathrm{C}\), what are the temperature and mass of the remaining ice when the tea and ice reach a common temperature? Neglect energy transfers with the environment.

Show that when the temperature of a liquid in a barometer changes by \(\Delta T\) and the pressure is constant, the liquid's height \(h\) changes by \(\Delta h=\beta h \Delta T\), where \(\beta\) is the coefficient of volume expansion. Neglect the expansion of the glass tube.

. Steam What mass of steam at \(100^{\circ} \mathrm{C}\) must be mixed with \(150 \mathrm{~g}\) of ice at its melting point, in a thermally insulated container, to produce liquid water at \(50^{\circ} \mathrm{C} ?\)

A tank of water has been outdoors in cold weather, and a slab of ice \(5.0 \mathrm{~cm}\) thick has formed on its surface (Fig. 19-42). The air above the ice is at \(-10^{\circ} \mathrm{C}\). Calculate the rate of formation of ice (in centimeters per hour) on the ice slab. Take the thermal conductivity and density of ice to be \(0.0040 \mathrm{cal} / \mathrm{s} \cdot \mathrm{cm} \cdot \mathrm{C}^{\circ}\) and \(0.92 \mathrm{~g} / \mathrm{cm}^{3}\). Assume that energy is not transferred through the walls or bottom of the tank.

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