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Chapter 12: Problem 83

Suppose you are selling apple cider for two dollars a gallon when the temperature is \(4.0^{\circ} \mathrm{C} .\) The coefficient of volume expansion of the cider is \(280 \times 10^{6}\left(\mathrm{C}^{\circ}\right)^{-1} .\) If the expansion of the container is ignored, how much more money (in pennies) would you make per gallon by refilling the container on a day when the temperature is \(26^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
You would make 1 penny more per gallon.

Step by step solution

01

Understand the Problem

We need to calculate how much additional apple cider can fit into the container when the temperature increases from \(4.0^{\circ} \mathrm{C}\) to \(26^{\circ} \mathrm{C}\). Then, we find out how much more money is earned from selling this extra cider.
02

Calculate Change in Temperature

The temperature change \(\Delta T\) is calculated as: \[ \Delta T = 26^{\circ} \mathrm{C} - 4.0^{\circ} \mathrm{C} = 22^{\circ} \mathrm{C} \]
03

Identify Volume Expansion Formula

The formula for volume expansion due to temperature change is: \[ \Delta V = V_0 \beta \Delta T \] where \(\Delta V\) is the change in volume, \(V_0\) is the original volume, \(\beta\) is the coefficient of volume expansion, and \(\Delta T\) is the change in temperature.
04

Apply Known Values

In our case, \(V_0 = 1\) gallon, \(\beta = 280 \times 10^{-6} \left(\mathrm{C}^{\circ}\right)^{-1}\), and \(\Delta T = 22^{\circ} \mathrm{C}\). Substitute these values into the volume expansion formula: \[ \Delta V = 1 \times 280 \times 10^{-6} \times 22 \]
05

Compute Volume Change

Perform the multiplication to find \(\Delta V\). \[ \Delta V = 280 \times 10^{-6} \times 22 = 0.00616 \text{ gallons} \]
06

Calculate Additional Revenue

You earn more money by selling the extra cider. At \(2\) dollars per gallon, calculate the additional revenue from \(0.00616\) gallons. \[ \text{Additional Revenue} = 0.00616 \times 2 \times 100 = 1.232 \text{ pennies} \]
07

Round to Whole Pennies

Since you can't have fractions of a penny, round \(1.232\) pennies to the nearest whole number. The final result is \(1\) penny.

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Key Concepts

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

Volume Expansion
Volume expansion is a key concept in understanding how materials change when temperatures vary. As the temperature of a substance rises, it often expands, leading to an increase in volume. This happens because the molecules within most materials move more vigorously and further apart as they absorb heat.
  • Volume expansion is expressed by the formula: \[ \Delta V = V_0 \beta \Delta T \]
  • \(\Delta V\) is the change in volume.
  • \(V_0\) represents the original volume.
  • \(\beta\) is the coefficient of volume expansion.
  • \(\Delta T\) represents the change in temperature.
This equation shows how changes in temperature, specifically increase, will lead to a volume expansion if the coefficient of expansion is greater than zero.
Temperature Change
Temperature change measures how much the temperature increases or decreases. In the exercise, we have to calculate how temperature variation will impact cider volume.
To compute this, you subtract the initial temperature from the final temperature. This is quantified using the formula: \[ \Delta T = T_{ ext{final}} - T_{ ext{initial}} \]
  • For instance, increasing from \(4.0^{\circ} \mathrm{C}\) to \(26^{\circ} \mathrm{C}\) results in a temperature change of \(22^{\circ} \mathrm{C}\).
Temperature change directly influences how much volume will expand, emphasizing its role as a critical factor in these calculations.
Coefficients of Expansion
The coefficient of expansion is a scalar that indicates how much a material expands per degree change in temperature. Every substance has its unique coefficient of expansion, which tells how sensitive it is to thermal expansion.
  • The specific coefficient of expansion relates back to volume changes in materials.
  • In our exercise, the cider has a coefficient of volume expansion of \(280 \times 10^{-6} (\mathrm{C}^{\circ})^{-1}\).
This coefficient explains that for each degree the temperature rises, the cider will expand by a minuscule amount per unit of original volume. Knowing the coefficient enables us to predict the extent of expansion for practical applications like pricing adjustments in this cider-selling scenario.

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Most popular questions from this chapter

Multiple-Concept Example 11 deals with a situation that is similar, but not identical, to that here. When \(4200 \mathrm{~J}\) of heat are added to a \(0.15-\mathrm{m}\) -long silver bar, its length increases by \(4.3 \times 10^{-3} \mathrm{~m}\). What is the mass of the bar?

At a fabrication plant, a hot metal forging has a mass of \(75 \mathrm{~kg}\) and a specific heat capacity of \(430 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) To harden it, the forging is immersed in \(710 \mathrm{~kg}\) of oil that has a temperature of \(32^{\circ} \mathrm{C}\) and a specific heat capacity of \(2700 \mathrm{~J} /\left(\mathrm{kg} \dot{\mathrm{c}} \mathrm{C}^{\circ}\right) .\) The final temperature of the oil and forging at thermal equilibrium is \(47^{\circ} \mathrm{C}\). Assuming that heat flows only between the forging and the oil, determine the initial temperature of the forging.

On the Rankine temperature scale, which is sometimes used in engineering applications, the ice point is at \(491.67^{\circ} \mathrm{R}\) and the steam point is at \(671.67^{\circ} \mathrm{R}\). Determine a relationship (analogous to Equation 12.1 ) between the Rankine and Fahrenheit temperature scales.

A steel bicycle wheel (without the rubber tire) is rotating freely with an angular speed of \(18.00 \mathrm{rad} / \mathrm{s}\). The temperature of the wheel changes from -100.0 to \(+300.0{ }^{\circ} \mathrm{C}\). No net external torque acts on the wheel, and the mass of the spokes is negligible. (a) Does the angular speed increase or decrease as the wheel heats up? Why? (b) What is the angular speed at the higher temperature?

The box of a well-known breakfast cereal states that one ounce of the cereal contains 110 Calories \((1\) food Calorie \(=4186 \mathrm{~J})\). If \(2.0 \%\) of this energy could be converted by a weight lifter's body into work done in lifting a barbell, what is the heaviest barbell that could be lifted a distance of \(2.1 \mathrm{~m}\) ?
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