/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Most automobiles have a coolant ... [FREE SOLUTION] | 91Ó°ÊÓ

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Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its \(16.0\) -L capacity when at \(10.0^{\circ} \mathrm{C}\). What volume of radiator fluid will overflow when the radiator and fluid reach their \(95.0^{\circ} \mathrm{C}\) operating temperature, given that the fluid's volume coefficient of expansion is \(\beta=400 \times 10^{6} /{ }^{\circ} \mathrm{C}\) ? Note that this coefficient is approximate, because most car radiators have operating temperatures of greater than \(95.0^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The volume of radiator fluid that will overflow is approximately \(0.544L\).

Step by step solution

01

Understanding Thermal Expansion

The volume expansion of liquids is given by the formula \(V = V_0(1 + \beta\Delta T)\), where \(V\) is the final volume after expansion, \(V_0\) is the initial volume before expansion, \(\beta\) is the volume coefficient of expansion, and \(\Delta T\) is the change in temperature.
02

Calculate Change in Temperature

First, determine the change in temperature by subtracting the initial temperature from the final operating temperature \(\Delta T = T_{final} - T_{initial} = 95.0^\circ C - 10.0^\circ C = 85.0^\circ C\).
03

Compute the Final Volume

Next, we use the initial volume \(V_0 = 16.0L\) and the given \(\beta = 400 \times 10^{-6} / { }^\circ C\) to compute the final volume using the formula \(V = V_0(1 + \beta\Delta T)\). So \(V = 16.0L(1 + 400 \times 10^{-6} / { }^\circ C \times 85.0^\circ C)\).
04

Determine the Volume of Overflow

Finally, subtract the initial volume from the final volume to determine the volume of radiator fluid that will overflow \(V_{overflow} = V - V_0\).

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Key Concepts

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

Volume Coefficient of Expansion
Understanding the volume coefficient of expansion is crucial for grasping how substances change volume when subjected to temperature variations. Imagine a sealed container full of a liquid. If the liquid's temperature rises, the particles inside move faster and tend to occupy a larger volume. This phenomenon is described quantitatively by the volume coefficient of expansion, denoted as \( \beta \).

It's a proportionality constant that reflects how much the volume of a substance changes per degree change in temperature. Specifically, the formula that connects these quantities is \( V = V_0(1 + \beta\Delta T) \), where \( V \) is the final volume, \( V_0 \) the initial volume, \( \beta \) the volume coefficient of expansion, and \( \Delta T \) the temperature change. For liquids, \( \beta \) is typically larger than for solids due to looser chemical bonds which allow more movement when heated.

To better visualize this coefficient in action, consider our radiator fluid example. Copper expands, but the liquid coolant expands even more due to its high \( \beta \). When calculating the overflow, this coefficient showed us precisely how much more volume the coolant will occupy as it heats up from \( 10.0^\circ \mathrm{C} \) to \( 95.0^\circ \mathrm{C} \).
Temperature Change in Physics
Temperature change plays a pivotal role in many physical phenomena, including thermal expansion. In physics, temperature not only measures the heat energy present in a substance but also relates directly to the average kinetic energy of the particles composing it. As temperature increases, particles move more vigorously, causing expansions in their respective substances.

In our radiator problem, the change in temperature (\( \Delta T \)) translates into the difference between the final and starting temperatures. Mathematically, it's expressed as \( \Delta T = T_{\text{final}} - T_{\text{initial}} \). It is this difference that, when applied to the volume coefficient of expansion formula, determines how much a substance expands or contracts.

Importance of Accurate Temperature Measurement

Getting the start and end temperatures right is critical as even small inaccuracies can lead to significant errors in calculating thermal expansion - especially for substances with high \( \beta \), like our radiator fluid. This sensitivity emphasizes the need for precise temperature controls in applications involving significant thermal expansion.
Thermal Expansion of Liquids
Liquids are particularly interesting when it comes to thermal expansion. Unlike solids, where atoms are locked into a crystal lattice, liquid particles have the freedom to move around more freely. This characteristic leads to more pronounced expansion in response to temperature increases.

When a liquid is heated, the increase in kinetic energy causes the particles to push against each other with greater force, leading to an increase in volume. However, unlike gases, liquids are relatively incompressible, so they won't expand indefinitely with pressure - a fact that must be considered in technical applications like cooling systems in cars.

Impact on Real-world Systems

Our automotive radiator example showcases the impact of thermal expansion on a real-world system. Engineers must design reservoirs to accommodate the increased volume of liquid to prevent damage or overflow. Otherwise, the expanding fluid can cause leaks or reduce the effectiveness of the cooling system. Remember, when a coolant's temperature reaches its operating limit, as in the given \( 95.0^\circ \mathrm{C} \) scenario, it's not just the volume of the fluid that increases - the radiator itself expands too (though not as much), further complicating the system's dynamics.

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Most popular questions from this chapter

A freight train consists of two \(8.00 \times 10^{4}\) -kg engines and 45 cars with average masses of \(5.50 \times 10^{4} \mathrm{kg}\). (a) What force must each engine exert backward on the track to accelerate the train at a rate of \(5.00 \times 10^{-2} \mathrm{m} / \mathrm{s}^{2}\) if the force of friction is \(7.50 \times 10^{5} \mathrm{N},\) assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy- efficient transportation systems. (b) What is the force in the coupling between the 37 th and 38 th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

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