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Chapter 17: Problem 83

A metal rod that is 30.0 \(\mathrm{cm}\) long expands by 0.0650 \(\mathrm{cm}\) when its temperature is raised from \(0.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} .\) A rod of a different metal and of the same length expands by 0.0350 \(\mathrm{cm}\) for the same rise in temperature. A third rod, also 30.0 \(\mathrm{cm}\) long, is made up of pieces of each of the above metals placed end to end and expands 0.0580 \(\mathrm{cm}\) between \(0.0^{\circ} \mathrm{C}\) and \(100.0^{\circ} \mathrm{C} .\) Find the length of each portion of the composite rod.

Short Answer

Expert verified
The composite rod consists of 26.0 cm of the first metal and 4.0 cm of the second metal.

Step by step solution

01

Understanding Thermal Expansion

Thermal expansion of a material can be described by the formula \( \Delta L = L \cdot \alpha \cdot \Delta T \), where \( \Delta L \) is the change in length, \( L \) is the initial length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.
02

Calculate Coefficients of Expansion

For the first metal rod: \( 0.0650 = 30.0 \cdot \alpha_1 \cdot 100 \). Solving for \( \alpha_1 \), we get \( \alpha_1 = \frac{0.0650}{30.0 \cdot 100} \). For the second metal rod: \( 0.0350 = 30.0 \cdot \alpha_2 \cdot 100 \), solving for \( \alpha_2 \), we get \( \alpha_2 = \frac{0.0350}{30.0 \cdot 100} \).
03

Calculate Expansion of Composite Rod

For the composite rod made from lengths \( x \) of the first metal and \( y \) of the second metal such that \( x + y = 30.0 \), the total expansion \( 0.0580 = x \cdot \alpha_1 \cdot 100 + y \cdot \alpha_2 \cdot 100 \).
04

Solve the System of Equations

From \( x + y = 30.0 \) and substituting \( y = 30 - x \) into the expansion equation gives \( 0.0580 = x \cdot \alpha_1 \cdot 100 + (30 - x) \cdot \alpha_2 \cdot 100 \). Substitute the values of \( \alpha_1 \) and \( \alpha_2 \) from Step 2 and solve for \( x \).
05

Calculate Lengths of Each Portion

Substituting back, we find \( x = 26.0 \text{ cm} \) and \( y = 4.0 \text{ cm} \). These represent the length of the first and second metal, respectively, in the composite rod.

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Key Concepts

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

Linear Expansion Coefficient
The linear expansion coefficient, denoted as \( \alpha \), is a fundamental property that quantifies how much a material expands per degree of temperature change. To calculate \( \alpha \) for any material, we use the formula: \[ \Delta L = L \cdot \alpha \cdot \Delta T \] where:
  • \( \Delta L \) is the change in length of the material.
  • \( L \) is the original length of the material before temperature change.
  • \( \Delta T \) is the change in temperature.
For instance, if we have a metal rod with an initial length of 30.0 cm that expands by 0.0650 cm when heated from 0掳C to 100掳C, the linear expansion coefficient can be calculated as: \[ \alpha = \frac{\Delta L}{L \cdot \Delta T} = \frac{0.0650}{30.0 \cdot 100} \]. This coefficient is specific to each material, effectively serving as a fingerprint for how a material responds to heat.
Composite Materials
Composite materials integrate multiple distinct materials to leverage the benefits of each. In the case of our rods, we have two different metals combined end to end, creating a composite rod. A unique challenge with composites is predicting how they behave under temperature changes due to the different linear expansion coefficients. For instance, if we have a composite rod made by joining two rods of different metals, it will expand by a combination of each metal's expansion:
  • The formula used is: \[ 0.0580 = x \cdot \alpha_1 \cdot 100 + y \cdot \alpha_2 \cdot 100 \] where \( x \) and \( y \) are lengths of each metal in the composite.
  • This accounts for the length contributions from each metal to the total expansion based on their specific \( \alpha \) values.
This composite approach helps in designing materials that exploit individual strengths while minimizing weaknesses, a common practice in engineering.
Temperature Change Effects
Temperature changes impact materials by causing them to expand or contract. When a metal is heated, its particles gain energy and move more, causing an increase in length. This property, while useful, requires careful consideration when designing materials needing precision, such as machinery components or engines. In the exercise, the metal rods expand differently when subject to the same temperature rise, illustrating the variance caused by their unique \( \alpha \) values.
  • For the metal rod with a higher \( \alpha \), a greater expansion is observed, while the rod with a lower \( \alpha \) expands less.
  • Temperature-induced expansion is crucial in scenarios like bridge construction where expansion joints are required to accommodate changes.
Understanding these effects helps in planning artifacts and structures to ensure they can withstand temperature fluctuations without structural failure.

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Most popular questions from this chapter

While running, a 70 -kg student generates thermal energy at a rate of 1200 \(\mathrm{W}\) . For the runner to maintain a constant body temperature of \(37^{\circ} \mathrm{C},\) this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the heat could not flow out of the student's body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to \(44^{\circ} \mathrm{C}\) or higher. The specific heat of a typical human body is \(3480 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K},\) slightly less than that of water. The difference is due to the presence of protein, fat,and minerals, which have lower specific heats.)

In a container of negligible mass, 0.0400 kg of steam at \(100^{\circ} \mathrm{C}\) and atmospheric pressure is added to 0.200 \(\mathrm{kg}\) of water at \(50.0^{\circ} \mathrm{C} .\) (a) If no heat is lost to the surroundings, what is the final temperature of the system? (b) At the final temperature, how many kilograms are there of steam and how many of liquid water?

\(\mathrm{A} 500.0-\mathrm{g}\) chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing 1.00 \(\mathrm{kg}\) of water at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of \(22.0^{\circ} \mathrm{C}\) (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) What if the heat absorbed by the Styrofoam actually is not negligible. How would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

(a) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.

A Walk in the Sun. Consider a poor lost soul walking at 5 \(\mathrm{km} / \mathrm{h}\) on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of \(280 \mathrm{W},\) and almost all of this energy is con- verted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k^{\prime} A_{\text { skin }}\left(T_{\text { air }}-T_{\text { skin }}\right),\) where \(k^{\prime}\) is \(54 \mathrm{J} / \mathrm{h} \cdot \mathrm{C}^{\circ} \cdot \mathrm{m}^{2},\) the exposed skin area \(A_{\text { skin }}\) is \(1.5 \mathrm{m}^{2},\) the air temperature \(T_{\mathrm{air}}\) is \(47^{\circ} \mathrm{C},\) and the skin temperature \(T_{\text { skin }}\) is \(36^{\circ} \mathrm{C} ;\) (iii) the skin absorbs radiant energy from the sun at a rate of 1400 \(\mathrm{W} / \mathrm{m}^{2}\) ; (iv) the skin absorbs radiant energy from the environment, which has temperature \(47^{\circ} \mathrm{C}\) . (a) Calculate the net rate (in watts) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e=1\) and that the skin temperature is initially \(36^{\circ} \mathrm{C}\) . Which mechanism is the most important? (b) At what rate (in \(\mathrm{L} / \mathrm{h} )\) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at \(36^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\)) (c) Suppose instead the person is protected by light-colored clothing \((e \approx 0)\) so that the exposed skin area is only 0.45 \(\mathrm{m}^{2} .\) What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.
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