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

Short Answer

Expert verified
The composite rod is 23.0 cm of the first metal and 7.0 cm of the second metal.

Step by step solution

01

Define the problem variables

Let the coefficient of linear expansion of the first metal be \( \alpha_1 \) and the length of this metal in the composite rod be \( x \) cm. Similarly, let the coefficient of linear expansion of the second metal be \( \alpha_2 \) and its length be \( y \) cm. Therefore, the composite rod's total length is \( x + y = 30.0 \) cm.
02

Use given expansions to find coefficient of expansion

For the first metal, the expansion is given by \( 0.0650 = \alpha_1 \cdot 30.0 \cdot 100 \). Thus, \( \alpha_1 = \frac{0.0650}{30.0 \times 100}\).For the second metal, the expansion is given by \( 0.0350 = \alpha_2 \cdot 30.0 \cdot 100 \). Thus, \( \alpha_2 = \frac{0.0350}{30.0 \times 100}\).
03

Set up the equation for the composite rod

The total expansion for the composite rod is given as \( 0.0580 \). Express the expansion of the composite rod as: \( \alpha_1 \cdot x \cdot 100 + \alpha_2 \cdot y \cdot 100 = 0.0580 \).
04

Substitute and simplify

Substitute \( \alpha_1 = \frac{0.0650}{3000} \) and \( \alpha_2 = \frac{0.0350}{3000} \) into the equation:\[\frac{0.0650}{3000} \cdot x \cdot 100 + \frac{0.0350}{3000} \cdot y \cdot 100 = 0.0580\]Simplifying, we get:\[0.065x + 0.035y = 1.740\]
05

Solve simultaneous equations

We now have the two equations:1. \( x + y = 30.0 \)2. \( 0.065x + 0.035y = 1.740 \)From equation 1, express \( y = 30.0 - x \). Substitute \( y \) in equation 2:\( 0.065x + 0.035(30.0 - x) = 1.740 \).Simplify to find \( x \):\[ 0.065x + 1.05 - 0.035x = 1.740 \]\[ 0.030x = 0.690 \]\[ x = 23.0 \]Now, substitute \( x = 23.0 \) into \( y = 30.0 - x \) to find \( y \):\( y = 30.0 - 23.0 = 7.0 \).
06

Verify the solution

The lengths of each portion are \( x = 23.0 \) cm and \( y = 7.0 \) cm. Check these values in both equations:1. \( x + y = 30.0 \) checks out.2. Substitute into possible second equation: \(0.065 \times 23.0 + 0.035 \times 7.0 = 1.495 + 0.245 = 1.740 \), which confirms correctness.

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Key Concepts

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

Coefficient of Linear Expansion
Materials expand when heated, and this expansion can be quantified using the coefficient of linear expansion. The coefficient of linear expansion, usually represented by the Greek letter \( \alpha \), describes how much a material expands per unit length per degree of temperature change. In mathematical terms, it is defined by the equation:
  • \( \Delta L = \alpha \cdot L_0 \cdot \Delta T \)
where:
  • \( \Delta L \) is the change in length,
  • \( \alpha \) is the coefficient of linear expansion,
  • \( L_0 \) is the original length,
  • \( \Delta T \) is the change in temperature.
This concept helps engineers and scientists predict how materials will behave in different environments, crucial for safety and functionality.
To calculate \( \alpha \) for different metals, given the original length and temperature change, rearrange the formula:
  • \( \alpha = \frac{\Delta L}{L_0 \cdot \Delta T} \)
Using this formula, we can find the expansion properties of each metal in various conditions.
Composite Materials
Composite materials consist of two or more distinct components which, when combined, produce a material with characteristics different from the individual components. This is beneficial when the desired properties for an application are not present in a single material.
When dealing with composite materials in thermal expansion, each component expands at different rates due to its unique coefficient of linear expansion. For example, in a composite rod where metals are combined, the thermal behavior can be complex but predictable when calculated separately for each segment.
The total expansion of a composite structure can be derived by summing the expansions of its individual components. This allows engineers to design materials that balance and optimize the performance of each part of the composite.
Composite rods, like the one in our problem, perfectly illustrate how understanding each section's properties and behavior can lead to effective material design and usage.
Length Calculation
Length calculation in the context of composite materials involves determining how much each part of the composite contributes to the total length, especially when affected by factors like temperature changes.
To determine the lengths of the individual parts of a composite rod, consider the relationship between their expansions. As shown in the exercise, one can set up equations based on the given expansions and the sum of the lengths to solve for unknowns.
  • We start by letting \( x \) and \( y \) represent the lengths of individual materials in the composite rod.
  • Using the equations \( x + y = 30.0 \) and the calculated expansions \( 0.065x + 0.035y = 1.740 \), we can solve simultaneous equations to find the specific lengths.
Such length calculations are critical for practical applications to ensure that the composite material will function correctly under various conditions. This ensures structures remain intact and operational in their intended environments.

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Most popular questions from this chapter

The African bombardier beetle (Stenaptinus insignis) can emit a jet of defensive spray from the movable tip of its abdomen (Fig. P17.91). The beetle's body has reservoirs containing two chemicals; when the beetle is disturbed, these chemicals combine in a reaction chamber, producing a compound that is warmed from 20\(^\circ\)C to 100\(^\circ\)C by the heat of reaction. The high pressure produced allows the compound to be sprayed out at speeds up to 19 m/s 168 km/h2, scaring away predators of all kinds. (The beetle shown in Fig. P17.91 is 2 cm long.) Calculate the heat of reaction of the two chemicals (in J/kg). Assume that the specific heat of the chemicals and of the spray is the same as that of water, \(4.19 \times 10{^3} J/kg \cdot K\), and that the initial temperature of the chemicals is 20\(^\circ\)C.

A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at 255\(^\circ\)C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

Careful measurements show that the specific heat of the solid phase depends on temperature (Fig. P17.117). How will the actual time needed for this cryoprotectant to come to equilibrium with the cold plate compare with the time predicted by using the values in the table? Assume that all values other than the specific heat (solid) are correct. The actual time (a) will be shorter; (b) will be longer; (c) will be the same; (d) depends on the density of the cryoprotectant.

One end of an insulated metal rod is maintained at 100.0\(^\circ\)C, and the other end is maintained at 0.00\(^\circ\)C by an ice-water mixture. The rod is 60.0 cm long and has a cross-sectional area of 1.25 cm\(^2\). The heat conducted by the rod melts 8.50 g of ice in 10.0 min. Find the thermal conductivity \(k\) of the metal.

A copper calorimeter can with mass 0.446 kg contains 0.0950 kg of ice. The system is initially at 0.0\(^\circ\)C. (a) If 0.0350 kg of steam at 100.0\(^\circ\)C and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

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