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A steel aircraft carrier is 370 m long when moving through the icy North Atlantic at a temperature of \(2.0^{\circ} \mathrm{C} .\) By how much does the carrier lengthen when it is traveling in the warm Mediterranean Sea at a temperature of \(21^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The carrier lengthens by approximately 0.0775 meters.

Step by step solution

01

Identify the Formula

To find out how much the carrier lengthens, we use the linear expansion formula: \( \Delta L = \alpha L_0 \Delta T \), where \( \alpha \) is the coefficient of linear expansion for steel, \( L_0 \) is the original length, and \( \Delta T \) is the change in temperature.
02

Gather Known Values

The original length \( L_0 \) is 370 m, the initial temperature is \( 2.0^{\circ} \mathrm{C} \), and the final temperature is \( 21^{\circ} \mathrm{C} \). The coefficient of linear expansion for steel \( \alpha \) is approximately \( 11 \times 10^{-6} / ^{\circ} \mathrm{C} \).
03

Calculate Temperature Change

Find the temperature change: \( \Delta T = 21^{\circ} \mathrm{C} - 2^{\circ} \mathrm{C} = 19^{\circ} \mathrm{C} \).
04

Apply the Formula

Substitute the values into the formula: \( \Delta L = (11 \times 10^{-6} / ^{\circ} \mathrm{C})(370 \text{ m})(19^{\circ} \mathrm{C}) \).
05

Compute the Length Change

Calculate \( \Delta L = 11 \times 10^{-6} \times 370 \times 19 \) to find the length change, which equals approximately 0.07751 m.

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Key Concepts

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

Linear Expansion Formula
When certain materials are heated, they tend to expand. This is especially important in situations involving metals. The linear expansion formula helps us calculate this change when dealing with linear (straight-line) dimensions. Its mathematical expression is given by \[ \Delta L = \alpha L_0 \Delta T \]. Here's a quick breakdown to clarify this formula:
  • \( \Delta L \) represents the change in length you want to find.
  • \( \alpha \) is the coefficient of linear expansion, which is specific to the material in question.
  • \( L_0 \) is the material's original or initial length.
  • \( \Delta T \) stands for the change in temperature.
All these components work together in a straightforward multiplication, showing how much a length will increase or decrease with temperature change. Understanding this formula is essential for calculating real-world expansion in structures or materials such as bridges, pipelines, and, in this case, an aircraft carrier.
Coefficient of Linear Expansion
The coefficient of linear expansion is a crucial factor in determining how much a material will expand or contract when a temperature change occurs. It quantifies the rate of expansion or contraction per degree change in temperature. Each material has its unique coefficient, reflecting its atomic structure and bonding.For steel, this value is typically around \( 11 \times 10^{-6} / ^{\circ} \mathrm{C} \). This means for every degree Celsius change in temperature, a meter of steel will change its length by about \( 11 \times 10^{-6} \) meters. Other materials have different coefficients, so always use the value that corresponds to your particular material.Having a grasp of this concept helps you predict how structures or objects react to thermal stress. By incorporating this density into the linear expansion formula, one can make accurate predictions about potential changes in size, which is invaluable for engineering and design.
Temperature Change
Temperature change, indicated as \( \Delta T \), is foundational in evaluating thermal expansion. It provides the measurable difference between the material's initial and final temperatures.In our exercise example, the initial temperature is \(2^{\circ} \mathrm{C} \) while the final is \(21^{\circ} \mathrm{C} \). The difference, calculated as \[ \Delta T = 21^{\circ} \mathrm{C} - 2^{\circ} \mathrm{C} = 19^{\circ} \mathrm{C} \], gives the exact range of temperature over which expansion might happen. This change is critical, as it directly impacts the length change of the material. Any miscalculation can lead to incorrect predictions of behavior under thermal conditions. Recognizing how to measure and apply temperature change in the context of thermal expansion helps ensure structures maintain their integrity across varying climates.

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Most popular questions from this chapter

The column of mercury in a barometer (see Figure 11.11 ) has a height of \(0.760 \mathrm{m}\) when the pressure is one atmosphere and the temperature is \(0.0^{\circ} \mathrm{C}\). Ignoring any change in the glass containing the mercury, what will be the height of the mercury column for the same one atmosphere of pressure when the temperature rises to \(38.0^{\circ} \mathrm{C}\) on a hot day?

A 42-kg block of ice at \(0^{\circ} \mathrm{C}\) is sliding on a horizontal surface. The initial speed of the ice is \(7.3 \mathrm{m} / \mathrm{s}\) and the final speed is \(3.5 \mathrm{m} / \mathrm{s}\). Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice. Determine the mass of ice that melts into water at \(0^{\circ} \mathrm{C}\).

Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. While a person is exercising, \(0.6 \mathrm{kg}\) of blood flows to the body's surface and releases \(2000 \mathrm{J}\) of energy. The blood arriving at the surface has the temperature of the body's interior, \(37.0^{\circ} \mathrm{C}\). Assuming that blood has the same specific heat capacity as water, determine the temperature of the blood that leaves the surface and returns to the interior.

Equal masses of two different liquids have the same temperature of \(25.0^{\circ} \mathrm{C} .\) Liquid \(\mathrm{A}\) has a freezing point of \(-68.0^{\circ} \mathrm{C}\) and a specific heat capacity of \(1850 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) Liquid \(\mathrm{B}\) has a freezing point of \(-96.0^{\circ} \mathrm{C}\) and a specific heat capacity of \(2670 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) The same amount of heat must be removed from each liquid in order to freeze it into a solid at its respective freezing point. Determine the difference \(L_{\mathrm{f}, \mathrm{A}}-L_{\mathrm{f}, \mathrm{B}}\) between the latent heats of fusion for these liquids.

An unknown material has a normal melting/freezing point of \(-25.0^{\circ} \mathrm{C},\) and the liquid phase has a specific heat capacity of \(160 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) One-tenth of a kilogram of the solid at \(-25.0^{\circ} \mathrm{C}\) is put into a \(0.150-\mathrm{kg}\) aluminum calorimeter cup that contains \(0.100 \mathrm{kg}\) of glycerin. The temperature of the cup and the glycerin is initially \(27.0^{\circ} \mathrm{C}\). All the unknown material melts, and the final temperature at equilibrium is \(20.0^{\circ} \mathrm{C} .\) The calorimeter neither loses energy to nor gains energy from the external environment. What is the latent heat of fusion of the unknown material?

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