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Chapter 14: Problem 17

\(\bullet\) The markings on an aluminum ruler and a brass ruler begin at the left end; when the rulers are at \(0.00^{\circ} \mathrm{C}\) , they are perfectly aligned. How far apart will the 20.0 \(\mathrm{cm}\) marks be on the two rulers at \(100.0^{\circ} \mathrm{C}\) if the left-hand ends are kept precisely aligned?

Short Answer

Expert verified
The marks will be 0.0080 cm apart at 100掳C.

Step by step solution

01

Define Thermal Expansion Equation

The change in length due to thermal expansion can be calculated using the formula: \( \Delta L = \alpha \times L_0 \times \Delta T \), where \( \Delta L \) is the change in length, \( \alpha \) is the coefficient of linear expansion, \( L_0 \) is the original length, and \( \Delta T \) is the change in temperature.
02

Identify Coefficients of Linear Expansion

For aluminum, the coefficient of linear expansion \( \alpha_a \) is approximately \( 23 \times 10^{-6} \; \text{掳C}^{-1} \). For brass, the coefficient \( \alpha_b \) is approximately \( 19 \times 10^{-6} \; \text{掳C}^{-1} \). These coefficients are needed for our calculations.
03

Calculate Change in Length for Aluminum

Apply the thermal expansion formula for the aluminum ruler: \( \Delta L_a = 23 \times 10^{-6} \; \text{掳C}^{-1} \times 20.0 \; \text{cm} \times (100.0 - 0.00) \; \text{掳C} = 0.0460 \; \text{cm} \). This is the expansion of the aluminum ruler.
04

Calculate Change in Length for Brass

Apply the thermal expansion formula for the brass ruler: \( \Delta L_b = 19 \times 10^{-6} \; \text{掳C}^{-1} \times 20.0 \; \text{cm} \times (100.0 - 0.00) \; \text{掳C} = 0.0380 \; \text{cm} \). This is the expansion of the brass ruler.
05

Determine Distance Between Ruler Marks at 100掳C

The distance between the 20.0 cm marks on the rulers at \( 100.0^{\circ} \mathrm{C} \) will be the difference in their expansions. Therefore, the distance is \( \Delta L_a - \Delta L_b = 0.0460 \; \text{cm} - 0.0380 \; \text{cm} = 0.0080 \; \text{cm} \).

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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 is a crucial concept when discussing thermal expansion. It represents how much a material expands per degree change in temperature. This property is specific to each material and influences how it behaves with heat.
The formula for thermal expansion is:
  • \( \Delta L = \alpha \times L_0 \times \Delta T \)
Where \( \Delta L \) is the change in length, \( \alpha \) is the coefficient of linear expansion, \( L_0 \) is the original length, and \( \Delta T \) is the temperature change. The larger the coefficient, the more a material will expand when heated. For instance, metals typically have a higher coefficient of linear expansion compared to non-metals.
Aluminum
Aluminum is a metal known for its light weight and excellent thermal conductivity. It also has a relatively high coefficient of linear expansion, making it prone to expanding more when heated.
The coefficient for aluminum is approximately \( 23 \times 10^{-6} \, \text{掳C}^{-1} \). This means that for a given temperature change, aluminum will expand in length more significantly compared to other materials with lower coefficients.
  • This property is particularly important in applications like construction materials and aircraft, where temperature changes are frequent.
  • It also means that precision tools made from aluminum must account for thermal expansion to ensure accuracy.
Brass
Brass, an alloy chiefly made of copper and zinc, is another material often used where some level of thermal expansion is acceptable. It has a lower coefficient of linear expansion compared to aluminum, which is approximately \( 19 \times 10^{-6} \, \text{掳C}^{-1} \).
Brass is valued for:
  • Its strength and corrosion resistance.
  • Applications that require some flexibility in expansion like pipes and musical instruments.
Understanding the thermal expansion of brass is critical especially in cases where it might be exposed to varying temperatures.
Change in Length
Change in length due to thermal expansion is an essential consideration in material design. As temperatures shift, materials expand or contract, which is calculated using the linear expansion formula. In the given exercise, we calculate this for both aluminum and brass over a temperature range from \(0.00^{\circ} \text{C}\) to \(100.0^{\circ} \text{C}\).
  • For aluminum, the length change is calculated as: \( \Delta L_a = 0.0460 \, \text{cm} \).
  • For brass, it's \( \Delta L_b = 0.0380 \, \text{cm} \).
  • The difference in these expansions allows us to find how far apart the marks on the rulers will be at \(100.0^{\circ} \text{C}\).
These calculations help in coordinating design and use strategies when working with different materials across temperature variations.

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Most popular questions from this chapter

\(\cdot\) A slab of a thermal insulator with a cross-sectional area of 100 \(\mathrm{cm}^{2}\) is 3.00 cm thick. Its thermal conductivity is 0.075 \(\mathrm{W} /(\mathrm{m} \cdot \mathrm{K}) .\) If the temperature difference between opposite faces is \(80 \mathrm{C}^{\circ},\) how much heat flows the slab in 1 day?

\(\cdot\) A metal rod is 40.125 \(\mathrm{cm}\) long at \(20.0^{\circ} \mathrm{C}\) and 40.148 \(\mathrm{cm}\) long at \(45.0^{\circ} \mathrm{C} .\) Calculate the average coefficient of linear expansion of the rod's material for this temperature range.

\(\bullet\) On-demand water heaters. Conventional hot-water heaters consist of a tank of water maintained at a fixed temperature. The hot water is to be used when needed. The drawback is that energy is wasted because the tank loses heat when it is not in use, and you can run out of hot water if you use too much. Some utility companies are encouraging the use of on- demand water heaters (also known as flash heaters), which consist of heating units to heat the water as you use it. No water tank is involved, so no heat is wasted. A typical household shower flow rate is 2.5 gal \(/ \min (9.46 \mathrm{L} / \mathrm{min}\) ) with the tap water being heated from \(50^{\circ} \mathrm{F}\left(10^{\circ} \mathrm{C}\right)\) to \(120^{\circ} \mathrm{F}\left(49^{\circ} \mathrm{C}\right)\) by the on-demand heater. What rate of heat input (either electrical or from gas) is required to operate such a unit, assuming that all the heat goes into the water?

\(\cdot\) Treatment for a stroke. One suggested treatment for a person who has suffered a stroke is to immerse the patient in an ice-water bath at \(0^{\circ} \mathrm{C}\) to lower the body temperature, which prevents damage to the brain. In one set of tests, patients were cooled until their internal temperature reached \(32.0^{\circ} \mathrm{C}\) . To treat a 70.0 \(\mathrm{kg}\) patient, what is the minimum amount of ice (at \(0^{\circ} \mathrm{C} )\) that you need in the bath so that its temperature remains at \(0^{\circ} \mathrm{C} ?\) The specific heat capacity of the human body is \(3480 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right),\) and recall that normal body temperature is \(37.0^{\circ} \mathrm{C}\) .

\(\cdot\) The Eiffel Tower in Paris is 984 ft tall and is made mostly of steel. If this is its height in winter when its temperature is \(-8.00^{\circ} \mathrm{C},\) how much additional vertical distance must you cover if you decide to climb it during a summer heat wave when its temperature is \(40.0^{\circ} \mathrm{C}\) ? (b) Express the coefficient of linear expansion of steel in terms of Fahrenheit degrees.
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