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Two solid objects are made of different materials. Their volumes and volume expansion coefficients are \(V_{1}\) and \(V_{2}\) and \(\beta_{1}\) and \(\beta_{2}\), respectively. It is observed that during a temperature change of \(\Delta T\), the volume of each object changes by the same amount. If \(V_{1}=2 V_{2}\) what is the ratio of the volume expansion coefficients?

Short Answer

Expert verified
Answer: The ratio of the volume expansion coefficients is 1/2.

Step by step solution

01

Recall the volume expansion formula

The formula for volume expansion, which gives the change in volume \(\Delta V\) due to a change in temperature \(\Delta T\), is: $$\Delta V = \beta V \Delta T$$ where \(\beta\) is the volume expansion coefficient, \(V\) is the initial volume, and \(\Delta T\) is the change in temperature.
02

Setup the equation for both solid objects

Since the problem states that the volume of each object changes by the same amount during the temperature change, we can write this relationship as: $$\Delta V_{1} = \Delta V_{2}$$ Now, substitute the volume expansion formula for each object into this equation: $$\beta_{1} V_{1} \Delta T = \beta_{2} V_{2} \Delta T$$
03

Cancel out common terms

First, notice that both sides of the equation have a \(\Delta T\) term, which we can cancel out: $$\beta_{1} V_{1} = \beta_{2} V_{2}$$
04

Substitute the given relation between the volumes

We're given that \(V_{1} = 2 V_{2}\). Substitute this into the equation: $$\beta_{1}(2V_{2}) = \beta_{2}V_{2}$$
05

Solve for the ratio of expansion coefficients

Now, we want to find the ratio \(\frac{\beta_{1}}{\beta_{2}}\). Divide both sides of the equation by \(2V_{2}\beta_{2}\): $$\frac{\beta_{1} (2 V_{2})}{2V_{2}\beta_{2}} = \frac{\beta_{2}V_{2}}{2V_{2}\beta_{2}}$$ Simplify: $$\frac{\beta_{1}}{\beta_{2}} = \frac{1}{2}$$ So, the ratio of the volume expansion coefficients is \(\frac{1}{2}\).

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Key Concepts

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

Thermal Expansion
Thermal expansion is a fundamental physical concept describing how the size of an object changes with a change in temperature. When materials are heated and their temperature increases, they typically expand in all three directions: length, width, and volume. This expansion occurs because the heat gives the particles within the material more energy, causing them to vibrate and move more. This increased movement needs more space, leading to an expansion. Thermal expansion affects solids, liquids, and gases, but the amount by which they expand can vary significantly.

Commonly, solids exhibit a linear expansion along lengths and a volumetric expansion for changes affecting their entire shape. The importance of understanding thermal expansion extends from everyday applications, like why metal lids become looser when heated, to engineering applications, where bridges and rails are designed with gaps to allow for the expansion and contraction due to temperature fluctuations. Simply put, almost everything around us changes size with temperature, and engineers and scientists use the principles of thermal expansion to anticipate and compensate for these changes.
Volume Expansion Formula
The volume expansion formula is a mathematical expression used to calculate how much an object's volume changes when its temperature changes. It is given by the equation:
\[ \Delta V = \beta V \Delta T \]
where \(\Delta V\) is the change in volume, \(\beta\) is the volume expansion coefficient, \(V\) is the initial volume, and \(\Delta T\) is the change in temperature. The volume expansion coefficient is a proportionality constant that characterizes how much a substance's volume will expand per degree of temperature change. Different materials will generally have different coefficients, as their structures respond uniquely to the input of thermal energy.

The beauty of the volume expansion formula lies in its simplicity. By just knowing the initial volume, the temperature change, and the material-specific volume expansion coefficient, one can predict the final volume. For example, when designing a heating system or a fuel tank, engineers would use this formula to ensure safety and functionality under temperature variations.
Temperature Change
A temperature change, denoted as \(\Delta T\), is the difference between the final temperature and the initial temperature of an object or a system. It is crucial in many physical processes and is the driver behind thermal expansion. Being able to accurately measure and control temperature changes is essential in a variety of fields, from the culinary arts to the operating rooms of hospitals, and in the operation of complex machinery.

Control of temperature is key in materials science as well; for instance, in handling substances which might expand or become less dense when heated. This characteristic fundamentally affects designs, safety standards, and operational procedures in engineering projects. The way materials respond to temperature changes is also pivotal in understanding and mitigating the effects of thermal expansion in the environment, particularly when significant temperature variations are experienced.

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Most popular questions from this chapter

The background temperature of the universe is a) \(6000 \mathrm{~K}\). b) \(288 \mathrm{~K}\). c) \(3 \mathrm{~K}\). d) \(2.73 \mathrm{~K}\). e) \(0 \mathrm{~K}\).

When a 50.0 -m-long metal pipe is heated from \(10.0^{\circ} \mathrm{C}\) to \(40.0^{\circ} \mathrm{C}\), it lengthens by \(2.85 \mathrm{~cm}\). a) Determine the linear expansion coefficient. b) What type of metal is the pipe made of?

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.

On a cool morning, with the temperature at \(15.0^{\circ} \mathrm{C}\), a painter fills a 5.00 -gal aluminum container to the brim with turpentine. When the temperature reaches \(27.0^{\circ} \mathrm{C}\), how much fluid spills out of the container? The volume expansion coefficient for this brand of turpentine is \(9.00 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\).

Two mercury-expansion thermometers have identical reservoirs and cylindrical tubes made of the same glass but of different diameters. Which of the two thermometers can be graduated to a better resolution? a) The thermometer with the smaller diameter tube will have better resolution. b) The thermometer with the larger diameter tube will have better resolution. c) The diameter of the tube is irrelevant; it is only the volume expansion coefficient of mercury that matters. d) Not enough information is given to tell.

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