A copper cube of side length \(40 . \mathrm{cm}\) is heated from \(20 .{ }^{\circ} \mathrm{C}\) to \(120{ }^{\circ} \mathrm{C}\). What is the change in the volume of the cube? The linear expansion coefficient of copper is \(17 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).

Short Answer

Expert verified
Answer: Approximately 326.4 cm³

Step by step solution

01

Find the initial volume of the cube

The initial volume V1 of the cube can be found using the formula \(V1 = a^3\), where 'a' is the side length. V1 = (40 cm)³ = 64000 cm³
02

Calculate the change in temperature

The change in temperature, ΔT, is the final temperature minus the initial temperature. ΔT = 120°C - 20°C = 100°C
03

Use the formula for volume expansion

We use the formula for the volume expansion due to temperature change: ΔV = \(V_1 × 3 × \alpha × \Delta{T}\), where ΔV is the change in volume, V1 is the initial volume, α is the linear expansion coefficient, and ΔT is the change in temperature.
04

Calculate the change in volume

Substitute the known values into the formula: ΔV = 64000 cm³ × 3 × (17 × 10⁻⁶ °C⁻¹) × 100°C ΔV ≈ 326.4 cm³ The change in the volume of the copper cube is approximately 326.4 cm³.

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Key Concepts

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

Volume Expansion
When materials are heated, they tend to expand. This property is crucial to understand in the context of physical behavior under varying temperatures. The expansion of material's volume when exposed to a temperature increase is known as volume expansion. Take the example of a copper cube: when it is heated, not only does each side of the cube get longer, but the entire volume of the cube also increases.

Volume expansion can be quantified using the relationship: \[\Delta V = V_1 \times \beta \times \Delta T\]where \(\Delta V\) is the change in volume, \(V_1\) is the initial volume, \(\beta\) is the volumetric expansion coefficient which for isotropic materials is typically 3 times the linear expansion coefficient \(\alpha\), and \(\Delta T\) is the change in temperature. In practical terms, if we take a material like copper with a reliable expansion coefficient, we can precisely predict how its volume will change with a given temperature increase, which is key to many industrial and engineering applications.
Linear Expansion Coefficient
At the heart of understanding volume expansion lies the concept of the linear expansion coefficient, commonly denoted by \(\alpha\). This coefficient is a measure of how much a material expands per degree change in temperature along any dimension. It’s a characteristic property of the material and depends on the material's composition and structure.

The formula for linear expansion is given by: \[\Delta L = L_1 \times \alpha \times \Delta T\]where \(\Delta L\) is the change in length, \(L_1\) is the initial length, and \(\Delta T\) is the change in temperature. In our example, the linear expansion coefficient \(\alpha\) for copper enables us to determine accurately how any dimension of a copper object will change with temperature. This is pivotal in designing systems where precision and tolerances are critical, such as in engineering components that must fit together exactly.
Temperature Change in Physics
Temperature change in physics is a core concept that refers to the variation in the degree of hotness or coldness of a body. In most physical scenarios, substances expand when heated and contract when cooled. The temperature change is the driving force behind these dimensional and volumetric transformations.

The change in temperature (\(\Delta T\)) is calculated as the final temperature minus the initial temperature. Understanding how temperature change impacts materials is crucial, especially in constructing and designing objects that must withstand variations in environmental temperature. This understanding helps engineers and designers to take into account the expected expansions or contractions to avoid structural failures or loss of functionality in temperature-sensitive devices.

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Most popular questions from this chapter

Would it be possible to have a temperature scale defined in such a way that the hotter an object or system got, the lower (less positive or more negative) its temperature was?

A plastic-epoxy sheet has uniform holes of radius \(1.99 \mathrm{~cm}\). The holes are intended to allow solid ball bear- ings with an outer radius of \(2.00 \mathrm{~cm}\) to just go through. Over what temperature rise must the plastic-epoxy sheet be heated so that the ball bearings will go through the holes? The linear expansion coefficient of plastic-epoxy is about \(1.3 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\).

The volume of \(1.00 \mathrm{~kg}\) of liquid water over the temperature range from \(0.00^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C}\) fits reasonably well to the polynomial function \(V=1.00016-\left(4.52 \cdot 10^{-5}\right) T+\) \(\left(5.68 \cdot 10^{-6}\right) T^{2}\), where the volume is measured in cubic meters and \(T\) is the temperature in degrees Celsius. a) Use this information to calculate the volume expansion coefficient for liquid water as a function of temperature. b) Evaluate your expression at \(20.0^{\circ} \mathrm{C}\), and compare the value to that listed in Table \(17.3 .\)

Some textbooks use the unit \(\mathrm{K}^{-1}\) rather than \({ }^{\circ} \mathrm{C}^{-1}\) for values of the linear expansion coefficient; see Table 17.2 How will the numerical values of the coefficient differ if expressed in \(\mathrm{K}^{-1}\) ?

In order to create a tight fit between two metal parts, machinists sometimes make the interior part larger than the hole into which it will fit and then either cool the interior part or heat the exterior part until they fogether. Suppose an aluminum rod with diameter \(D_{1}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right)\) is to be fit into a hole in a brass plate that has a diameter \(D_{2}=10.000 \mathrm{~mm}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right) .\) The machinists can cool the rod to \(77.0 \mathrm{~K}\) by immersing it in liquid nitrogen. What is the largest possible diameter that the rod can have at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\) and just fit into the hole if the rod is cooled to \(77.0 \mathrm{~K}\) and the brass plate is left at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C} ?\) The linear expansion coefficients for aluminum and brass are \(22 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\) and \(19 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\), respectively.

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