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Chapter 19: Problem 36

An aluminum-alloy rod has a length of \(10.000 \mathrm{~cm}\) at \(20.000^{\circ} \mathrm{C}\) and a length of \(10.015 \mathrm{~cm}\) at the boiling point of water. (a) What is the length of the rod at the freezing point of water? (b) What is the temperature if the length of the rod is \(10.009 \mathrm{~cm}\) ?

Short Answer

Expert verified
a) 9.9981 cm b) 68掳C

Step by step solution

01

Determine the linear expansion coefficient

First, use the formula for linear expansion: \[ L = L_0(1 + \beta \triangle T) \]where \( L \) is the final length, \( L_0 \) is the initial length, \( \beta \) is the coefficient of linear expansion, and \( \triangle T \) is the change in temperature. Given lengths at specific temperatures, we can find \( \beta \). For the boiling point of water (\(100^{\text{\circ}} C\)) and initial conditions \(20^{\text{\circ}} C\), the equation becomes:\[ 10.015 \text{ cm} = 10.000 \text{ cm} (1 + \beta \times 80 \text{ C}) \]Solving for \( \beta \), we find:\[ \beta = \frac{10.015 \text{ cm} - 10.000 \text{ cm}}{10.000 \text{ cm} \times 80 \text{ C}} = 1.875 \times 10^{-5} \text{ C}^{-1} \]
02

Calculate the length at the freezing point

For the freezing point of water (\(0^{\text{\circ}} C\)), use the linear expansion formula again:\[ L = 10.000 \text{ cm} (1 + 1.875 \times 10^{-5} \text{ C}^{-1} \times (0^{\text{\circ}} C - 20^{\text{\circ}} C)) \]This simplifies to:\[ L = 10.000 \text{ cm} (1 - 1.875 \times 10^{-4}) = 10.000 \text{ cm} - 0.001875 \text{ cm} = 9.998125 \text{ cm} \]
03

Calculate the temperature for a rod length of 10.009 cm

Use the same linear expansion formula to solve for temperature when the rod length is 10.009 cm:\[ 10.009 \text{ cm} = 10.000 \text{ cm} (1 + 1.875 \times 10^{-5} \text{ C}^{-1} \times (T - 20^{\text{\circ}} C)) \]Simplify and solve for \( T \):\[ \frac{10.009 \text{ cm}}{10.000 \text{ cm}} = 1 + 1.875 \times 10^{-5} \text{ C}^{-1} \times (T - 20^{\text{\circ}} C) \]\[ 1.0009 = 1 + 1.875 \times 10^{-5} \text{ C}^{-1} \times (T - 20^{\text{\circ}} C) \]\[ 0.0009 = 1.875 \times 10^{-5} \text{ C}^{-1} \times (T - 20^{\text{\circ}} C) \]\[ \frac{0.0009}{1.875 \times 10^{-5}} = T - 20 \]\[ T - 20 = 48 \]\[ T = 68^{\text{\circ}} C \]

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Key Concepts

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

coefficient of linear expansion
The coefficient of linear expansion, denoted as \( \beta \), is a material-specific value that describes how much a material's length will change per degree change in temperature.
In the context of the exercise, we use the formula:
\( L = L_0(1 + \beta \, \Delta T) \)
Here, \( L \) is the final length, \( L_0 \) is the initial length, and \( \Delta T \) represents the change in temperature.
Essentially, this formula lets us calculate the material's new length when the temperature changes, considering its linear expansion coefficient.
In the exercise, we determined \( \beta \) for the aluminum-alloy rod by utilizing given lengths at specified temperatures. This involved straightforward algebra to isolate and solve for \( \beta \).
The determined coefficient for the aluminum-alloy is:
\( \beta = 1.875 \times 10^{-5} \, \text{C}^{-1} \).
temperature change
Temperature change, denoted as \( \Delta T \), is the difference between the final and initial temperatures. In simpler terms, it tells us how much the temperature has changed during a process.
In our exercise, we had three key temperatures:
  • The initial 20掳C
  • The boiling point of water 100掳C
  • The freezing point of water 0掳C

For the boiling point of water:
\( \Delta T = 100掳C - 20掳C = 80掳C \)
For calculating rod length at 0掳C:
\( \Delta T = 0掳C - 20掳C = -20掳C \)
Changes in temperature directly impact the rod's length as described by the linear expansion formula. By adjusting \( \Delta T \), we derive the rod's length at new temperatures.
initial and final lengths
Initial and final lengths are crucial in understanding and calculating linear expansion.
The **initial length** (denoted as \( L_0 \)) is the original measure of the rod before any temperature change.
In the exercise, the initial length of the rod at 20掳C is:
\( L_0 = 10.000 \, \text{cm} \)
The **final length** (denoted as \( L \)) is the length of the rod after the temperature change.
For example, the rod's length at the boiling point of water (100掳C) is:
\( L = 10.015 \, \text{cm} \)
Using the formula for linear expansion, we can find final lengths at different temperatures.
For freezing point, calculated as:
\( L = 10.000 \, \text{cm} (1 - 1.875 \times 10^{-4}) = 9.998125 \, \text{cm} \)
For a rod length of 10.009 cm, determining the temperature:
\( 10.009 \, \text{cm} = 10.000 \, \text{cm} (1 + 1.875 \times 10^{-5} \, \text{C}^{-1} \times (T - 20)) \)
Solving for \( T \), we find: \( T = 68掳C \)

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Most popular questions from this chapter

A person makes a quantity of iced tea by mixing 500 \(\mathrm{g}\) of hot tea (essentially water) with an equal mass of ice at its melting point. If the initial hot tea is at a temperature of (a) \(90^{\circ} \mathrm{C}\) and (b) \(70^{\circ} \mathrm{C}\), what are the temperature and mass of the remaining ice when the tea and ice reach a common temperature? Neglect energy transfers with the environment.

alcohol has a boiling point of \(78^{\circ} \mathrm{C}\), a freezing point of \(-114^{\circ} \mathrm{C}\), a heat of vaporization of \(879 \mathrm{~kJ} / \mathrm{kg}\), a heat of fusion of \(109 \mathrm{~kJ} / \mathrm{kg}\), and a specific heat of \(2.43 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{C}^{\circ}\). How much thermal energy must be transferred out of \(0.510 \mathrm{~kg}\) of ethyl alcohol that is initially a gas at \(78^{\circ} \mathrm{C}\) so that it becomes a solid at \(-114^{\circ} \mathrm{C} ?\)

(a) What is the rate of thermal energy transfer in watts per square meter through a glass window \(3.0 \mathrm{~mm}\) thick if the outside temperature is \(-20^{\circ} \mathrm{F}\) and the inside temperature is \(+72^{\circ} \mathrm{F} ?\) (b) A storm window having the same thickness of glass is installed parallel to the first window, with an air gap of \(7.5 \mathrm{~cm}\) between the two windows. What now is the rate of energy loss if conduction is the only important energytransfer mechanism?

(a) Calculate the rate at which body heat is conducted through the clothing of a skier in a steady-state process, given the following data: the body surface area is \(1.8 \mathrm{~m}^{2}\) and the clothing is \(1.0 \mathrm{~cm}\) thick; the skin surface temperature is \(33^{\circ} \mathrm{C}\) and the outer surface of the clothing is at \(1.0^{\circ} \mathrm{C} ;\) the thermal conductivity of the clothing is \(0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (b) How would the answer to (a) change if, after a fall, the skier's clothes became soaked with water of thermal conductivity \(0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} ?\)

A room is lighted by four \(100 \mathrm{~W}\) incandescent lightbulbs. (The power of \(100 \mathrm{~W}\) is the rate at which a bulb converts electrical energy to thermal energy and the energy of visible light.) Assuming that \(90 \%\) of the energy is converted to thermal energy, how much thermal energy is transferred to the room in \(1.00 \mathrm{~h}\) ?
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