/*! 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 25 A plastic container is completel... [FREE SOLUTION] | 91Ó°ÊÓ

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A plastic container is completely filled with gasoline at \(15^{\circ} \mathrm{C}\). The container specifications indicate that it can endure a \(1 \%\) increase in volume before rupturing. Within the temperatures likely to be experienced by the gasoline, the average coefficient of volume expansion of gasoline is \(9.5 \times 10^{-4} \mathrm{~K}^{-1}\). Estimate the maximum temperature rise that can be endured by the gasoline without causing a rupture in the storage container.

Short Answer

Expert verified
The maximum temperature rise is about \(10.53^{ ext{o}} ext{C}\).

Step by step solution

01

Understand the Problem

We need to find the maximum temperature rise possible for gasoline in a container, without causing the container to rupture. The container ruptures if the gasoline expands more than 1% of its volume.
02

Use the Volume Expansion Formula

The formula to calculate the volume expansion due to temperature increase is given by the equation: \( \Delta V = \beta V_0 \Delta T \), where \( \Delta V \) is the change in volume, \( \beta \) is the coefficient of volume expansion, \( V_0 \) is the initial volume, and \( \Delta T \) is the temperature change to be determined.
03

Relate Volume Expansion to Maximum Tolerable Increase

Since the container can withstand a 1% increase in volume: \( \Delta V = 0.01 V_0 \). Thus, \( 0.01 V_0 = \beta V_0 \Delta T \). This simplifies to: \( \Delta T = \frac{0.01}{\beta} \).
04

Substituting Values

Substitute the given value for the coefficient of volume expansion, \( \beta = 9.5 \times 10^{-4} \mathrm{~K}^{-1} \), into the formula: \( \Delta T = \frac{0.01}{9.5 \times 10^{-4}} \).
05

Calculate Maximum Temperature Rise

Perform the arithmetic: \( \Delta T = \frac{0.01}{9.5 \times 10^{-4}} \approx 10.53 \mathrm{~K} \). Thus, the maximum temperature rise that can be endured by the gasoline without rupturing the container is approximately \(10.53^{ ext{o}} ext{C}\).

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Key Concepts

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

Coefficient of Volume Expansion
The coefficient of volume expansion is a crucial concept for understanding how substances change in volume with temperature. This coefficient, often denoted as \( \beta \), measures how much a material expands per degree change in temperature. It is expressed in units of \( \mathrm{K}^{-1} \) or \( \mathrm{C}^{-1} \).When a liquid like gasoline heats up, its molecules move more vigorously, leading to an increase in volume. The degree to which the volume changes is proportional to both the temperature change and the coefficient of volume expansion.
  • A high \( \beta \) value indicates that the substance will expand significantly even with a small rise in temperature.
  • A low \( \beta \) implies that temperature changes have less impact on the material's volume.
For gasoline, \( \beta = 9.5 \times 10^{-4} \mathrm{~K}^{-1} \), meaning it has a relatively high capacity to expand with temperature changes compared to many other liquids.
Temperature Rise Calculation
Calculating the temperature rise that a substance can endure involves applying the formula for volume expansion. When a material like gasoline is contained, any extra expansion due to heat can lead to significant pressure build-up, potentially causing a rupture.To find the allowable increase in temperature \( ( \Delta T ) \), you use:\[\Delta T = \frac{\Delta V}{\beta V_0}\]Where:
  • \( \Delta V \) is the change in volume.
  • \( \beta \) is the coefficient of volume expansion of the liquid.
  • \( V_0 \) is the initial volume of the liquid.
In the case of the gasoline exercise, the maximum tolerable increase in volume (before the container ruptures) is 1% of the container's volume. Thus, by rearranging and simplifying the formula:\[\Delta T = \frac{0.01}{9.5 \times 10^{-4}} \approx 10.53 \mathrm{~K}\]This means the gasoline can safely endure a temperature rise of approximately 10.53°C without risking the container's integrity.
Gasoline Expansion
Gasoline expansion is an important consideration in storage and transport. As a fluid, gasoline is subject to changes in volume based on temperature variance, largely due to its relatively high coefficient of volume expansion. In practical scenarios, when gasoline's temperature increases, it expands, potentially leading to overflow or, in enclosed containers, pressure build-up. This is especially important in vehicles and storage units where temperature fluctuations are common.
  • In colder climates, as gasoline warms from cold to a more moderate temperature, it may occupy significantly more volume.
  • Similarly, in hot environments, the expansion becomes a critical factor in storage design and safety.
Proper design of storage containers must account for this expansion to prevent the risks associated with rupture. In the example, a container can withstand up to a 1% expansion, translating to a temperature rise of about 10.53°C, making understanding these principles essential for safe handling and storage.

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Most popular questions from this chapter

An equation that is commonly used to describe the volume flow rate of water in an open channel is given by $$ Q=\frac{1}{n} \frac{A^{\frac{5}{3}}}{P^{\frac{2}{3}}} S_{0}^{\frac{1}{2}} $$ where \(Q\) is the volume flow rate in the channel \(\left[\mathrm{L}^{3} \mathrm{~T}^{-1}\right], n\) is a constant that characterizes the roughness of the channel surface [dimensionless], \(A\) is the flow area \(\left[\mathrm{L}^{2}\right]\), \(P\) is the perimeter of the flow area that is in contact with the channel boundary [L], and \(S_{0}\) is the slope of the channel [dimensionless]. This equation is usually applied using SI units, where \(Q\) is in \(\mathrm{m}^{3} / \mathrm{s}, A\) is in \(\mathrm{m}^{2},\) and \(P\) is in \(\mathrm{m}\). (a) Is the given equation dimensionally homogeneous? (b) If the equation is not dimensionally homogeneous, what conversion factor must be inserted after the equal sign for the equation to work with \(Q\) in \(\mathrm{ft}^{3} / \mathrm{s}, A\) in \(\mathrm{ft}^{2},\) and \(P\) in \(\mathrm{ft}\) ?

Use prefixes to express the following quantities with magnitudes in the range of \(0.01-1000:(\) a \() 6.27 \times 10^{7} \mathrm{~N},\) (b) \(7.28 \times 10^{5} \mathrm{~Pa},\) (c) \(4.76 \times 10^{-4} \mathrm{~m}^{2},\) and (d) \(8.56 \times\) \(10^{5} \mathrm{~m}\).

A 1 -mm-diameter glass capillary tube is inserted in a beaker of mercury at \(20^{\circ} \mathrm{C}\). Previous experimenters report that the contact angle between mercury and the glass material is \(127^{\circ} .\) What is the expected depth of depression of mercury in the capillary tube?

A liquid with a surface tension of \(0.072 \mathrm{~N} / \mathrm{m}\) is used to form 60 -mm-diameter bubbles in air. What is the difference between the air pressure inside the bubble and the air pressure outside the bubble?

Before embarking on a 40-minute drive on the highway, a motorist adjusts his tire pressure to \(207 \mathrm{kPa}\). At the end of his trip, the motorist measures his tire pressure as \(241 \mathrm{kPa}\). Assume that the volume of the tire remains constant during the trip. (a) Estimate the percentage increase in the temperature of the air in the tire. (b) If the initial temperature of the air in the tire is assumed to be equal to the ambient air temperature of \(25^{\circ} \mathrm{C},\) what is the estimated temperature of the air in the tire at the end of the trip.

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