Problem 32
Question
A bar of iron is \(10 \mathrm{~cm}\) at \(20^{\circ} \mathrm{C}\). At \(19^{\circ} \mathrm{C}\) it will be \((\alpha\) of iron \(\left.=11 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)\) (a) \(11 \times 10^{-6} \mathrm{~cm}\) longer (b) \(11 \times 10^{-6} \mathrm{~cm}\) shorter (c) \(11 \times 10^{-5} \mathrm{~cm}\) shorter (d) \(11 \times 10^{-5} \mathrm{~cm}\) longer
Step-by-Step Solution
Verified Answer
The iron bar will be 11 \times 10^{-5} cm shorter, so option (c) is correct.
1Step 1: Understand the Concept of Thermal Expansion
Thermal expansion refers to the tendency of matter to change in shape, area, and volume in response to a change in temperature. The change in length, \(\Delta L\), of a material with an initial length of \(L_0\) and a temperature change \(\Delta T\) is calculated using the formula: \[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \] where \(\alpha\) is the coefficient of linear expansion.
2Step 2: Identify Known Values
Given a bar of iron: \(L_0 = 10\ \text{cm}\), \(\alpha = 11 \times 10^{-6}\ /{}^{\circ}C\), and temperature changes from \(20^{\circ}C\) to \(19^{\circ}C\). Thus, \(\Delta T = 19^{\circ}C - 20^{\circ}C = -1^{\circ}C\).
3Step 3: Apply the Formula for Change in Length
Substitute the known values into the expansion formula: \[ \Delta L = 11 \times 10^{-6} /{}^{\circ}C \times 10\ \text{cm} \times -1^{\circ}C \] Simplifying this, we find: \[ \Delta L = -1.1 \times 10^{-5}\ \text{cm} \]
4Step 4: Determine if Change is Expansion or Contraction
The negative sign in \(\Delta L = -1.1 \times 10^{-5}\ \text{cm}\) indicates the bar of iron has contracted (shortened) due to the temperature decrease.
5Step 5: Compare with Given Options
The calculated change in length \(-1.1 \times 10^{-5}\ \text{cm}\) corresponds to option (c) which states \(11 \times 10^{-5}\ \text{cm}\) shorter.
Key Concepts
Coefficient of Linear ExpansionTemperature ChangeLength Contraction
Coefficient of Linear Expansion
The coefficient of linear expansion, represented as \( \alpha \), is a crucial property that describes how the dimensions of a material change with temperature. It quantifies the fractional change in length per degree of temperature change. The formula for linear expansion is given by:\[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \]- \( \Delta L \) is the change in length.- \( \alpha \) is the coefficient of linear expansion.- \( L_0 \) is the original length of the material.- \( \Delta T \) is the temperature change.Each material has a unique \( \alpha \), dependent on its composition and structure. A higher value of \( \alpha \) signifies that the material expands more with an increase in temperature. It is an innate characteristic that helps engineers and scientists predict how materials behave under temperature variations.
This property ensures that structures and components can operate safely and effectively without unexpected failures due to temperature changes. For example, railway tracks and bridges are designed considering the coefficient of linear expansion to avoid bending or breaking.
This property ensures that structures and components can operate safely and effectively without unexpected failures due to temperature changes. For example, railway tracks and bridges are designed considering the coefficient of linear expansion to avoid bending or breaking.
Temperature Change
Temperature change, denoted as \( \Delta T \), represents the difference in temperature that a material experiences. It directly affects the physical properties of the material, particularly its dimensions.
When a material heats up, its particles move more vigorously, causing the material to expand. Conversely, a temperature decrease results in contraction as particle motion is reduced. The equation used for thermal expansion utilizes \( \Delta T \) to determine how a material will change in length given this temperature change:\[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \]In this context:- An increase in temperature (positive \( \Delta T \)) leads to expansion.- A decrease in temperature (negative \( \Delta T \)) results in contraction.
This concept is vital in practical applications, such as ensuring that pipes do not burst in freezing conditions and maintaining the integrity of electronic components under varying temperatures. Accurate predictions of length changes due to temperature shifts can prevent mechanical failures.
When a material heats up, its particles move more vigorously, causing the material to expand. Conversely, a temperature decrease results in contraction as particle motion is reduced. The equation used for thermal expansion utilizes \( \Delta T \) to determine how a material will change in length given this temperature change:\[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \]In this context:- An increase in temperature (positive \( \Delta T \)) leads to expansion.- A decrease in temperature (negative \( \Delta T \)) results in contraction.
This concept is vital in practical applications, such as ensuring that pipes do not burst in freezing conditions and maintaining the integrity of electronic components under varying temperatures. Accurate predictions of length changes due to temperature shifts can prevent mechanical failures.
Length Contraction
Length contraction refers to the shortening of a material's dimensions when the temperature decreases. As opposed to expansion, contraction occurs when the thermal energy decreases, causing the material's particles to settle into a tighter formation.
In the context of thermal expansion formulas, if \( \Delta T \) is negative, indicating a decrease in temperature, the calculated \( \Delta L \) will also be negative, representing length contraction:\[ \Delta L = \alpha \cdot L_0 \cdot (-\Delta T) \]For example, if an iron rod has an \( L_0 \) of 10 cm with a \( \Delta T \) of \(-1^{\circ}C\), and an \( \alpha \) of \( 11 \times 10^{-6} /^{\circ}C \), the rod contracts by \( -1.1 \times 10^{-5} \) cm. Understanding contraction is essential for:
In the context of thermal expansion formulas, if \( \Delta T \) is negative, indicating a decrease in temperature, the calculated \( \Delta L \) will also be negative, representing length contraction:\[ \Delta L = \alpha \cdot L_0 \cdot (-\Delta T) \]For example, if an iron rod has an \( L_0 \) of 10 cm with a \( \Delta T \) of \(-1^{\circ}C\), and an \( \alpha \) of \( 11 \times 10^{-6} /^{\circ}C \), the rod contracts by \( -1.1 \times 10^{-5} \) cm. Understanding contraction is essential for:
- Designing buildings and machinery that withstand cold temperatures without structural damage.
- Precision engineering, such as in aerospace or electronics, where even minor changes in dimension can have significant consequences.
- Developing proper material composites that mitigate contraction through adjusted material properties.
Other exercises in this chapter
Problem 30
The power radiated by a black body is \(P\), and it radiates maximum energy around the wavelength \(\lambda_{0}\). If the temperature of black body is now chang
View solution Problem 32
\(22 \mathrm{~g}\) of carbon dioxide at \(27^{\circ} \mathrm{C}\) is mixed in a closed container with \(16 \mathrm{~g}\) of oxygen at \(37^{\circ} \mathrm{C}\).
View solution Problem 33
A bar of iron is \(10 \mathrm{~cm}\) at \(20^{\circ} \mathrm{C}\). At \(19^{\circ} \mathrm{C}\) it will be \((\alpha\) of iron \(\left.=11 \times 10^{-6} /{ }^{
View solution Problem 34
The volume of a metal sphere increases by \(0.24 \%\) when its temperature is raised by \(40^{\circ} \mathrm{C}\). The coefficient of linear expansion of the me
View solution