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\(\cdot\) 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.

Short Answer

Expert verified
The coefficient of linear expansion is approximately \(2.294 \times 10^{-5} \, \mathrm{per} \, ^{\circ} \mathrm{C}\).

Step by step solution

01

Identify Given Values

First, write down the known information:- Initial length of the rod, \(L_0 = 40.125\, \mathrm{cm}\)- Final length of the rod, \(L = 40.148\, \mathrm{cm}\)- Initial temperature, \(T_0 = 20.0^{\circ} \mathrm{C}\)- Final temperature, \(T = 45.0^{\circ} \mathrm{C}\)
02

Understand the Formula

The formula for linear 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 initial length, and \(\Delta T\) is the change in temperature. We need to solve for \(\alpha\).
03

Calculate Change in Length

Calculate the change in length, \(\Delta L\):\[ \Delta L = L - L_0 = 40.148\, \mathrm{cm} - 40.125\, \mathrm{cm} = 0.023\, \mathrm{cm}\]
04

Calculate Change in Temperature

Find the change in temperature, \(\Delta T\):\[ \Delta T = T - T_0 = 45.0^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = 25.0^{\circ} \mathrm{C}\]
05

Solve for Coefficient of Linear Expansion

Rearrange the formula to solve for \(\alpha\):\[\alpha = \frac{\Delta L}{L_0 \times \Delta T}\]Substitute in the known values:\[\alpha = \frac{0.023\, \mathrm{cm}}{40.125\, \mathrm{cm} \times 25.0^{\circ} \mathrm{C}} = \frac{0.023}{1003.125}\]Calculate the value:\[\alpha \approx 2.294 \times 10^{-5} \, \mathrm{per} \, ^{\circ} \mathrm{C}\]
06

Conclusion

The average coefficient of linear expansion of the rod's material for this temperature range is approximately \(2.294 \times 10^{-5} \, \mathrm{per} \, ^{\circ} \mathrm{C}\).

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Key Concepts

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

Linear Expansion
Linear expansion refers to the increase in length of a material as it becomes warmer. In simpler terms, when materials, such as metal, heat up, they expand slightly and become longer. This is a common occurrence in everyday life and is especially significant in engineering and construction.
The change in length, denoted as \( \Delta L \), is calculated using the formula \( \Delta L = \alpha \times L_0 \times \Delta T \). Here, \( L_0 \) is the initial length of the material, \( \Delta T \) is the change in temperature, and \( \alpha \) is the coefficient of linear expansion. \( \alpha \) is unique for each material and indicates how much a material will expand per degree of temperature change.

Understanding linear expansion is crucial because it helps us predict how much a material will expand when subjected to heat. Engineers use this concept to design structures that can accommodate expansion and contraction without damage.
Thermal Expansion
Thermal expansion is a broad concept that includes not only linear expansion but also volumetric and area expansion. It is the tendency of matter to change its shape, area, volume, and density in response to a change in temperature.

Expanding on the idea of linear expansion, thermal expansion is critical in different realms:
  • Volumetric Expansion: This occurs when the entire volume of a solid, liquid, or gas expands with temperature increase. It affects how substances like liquids in a closed container behave when heated.
  • Area Expansion: Similar to linear expansion but applied to two-dimensional areas. It's less common in practical applications than linear or volumetric expansion but still helpful in some engineering calculations.
When engineers and scientists talk about thermal expansion, they often consider these factors holistically to ensure safety and functionality in various applications, from building bridges to developing everyday appliances.
Temperature Change
Temperature change is the difference between the initial and final temperatures in any expansion process. This change is a vital part of the linear expansion and thermal expansion calculation as it drives the extent to which materials expand or contract.
In practical scenarios, knowing the temperature change helps in:
  • Predicting Material Behavior: Helps in determining how a certain temperature increase or decrease will affect materials used in construction or manufacturing.
  • Designing Structures: Assists engineers in designing buildings, railways, and roads that withstand temperature fluctuations without significant damage.
In the context of the exercise given, the temperature change \( \Delta T = 45.0^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = 25.0^{\circ} \mathrm{C} \) is crucial for calculating how much the metal rod will expand when the temperature rises from 20°C to 45°C.

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Most popular questions from this chapter

. Hot air in a physics lecture. (a) A typical student listening attentively to a physics lecture has a heat output of 100 W. How much heat energy does a class of 90 physics students release into a lecture hall over the course of a 50 min lecture? (b) Assume that all the heat energy in part (a) is transferred to the 3200 \(\mathrm{m}^{3}\) of air in the room. The air has a specific heat capacity of 1020 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\) and a density of 1.20 \(\mathrm{kg} / \mathrm{m}^{3} .\) If none of the heat escapes and the air- conditioning system is off, how much will the temperature of the air in the room rise during the 50 min lecture? (c) If the class is taking an exam, the heat output per student rises to 280 \(\mathrm{W}\) . What is the temperature rise during 50 min in this case?

\(\bullet\) (a) On January \(22,1943,\) the temperature in Spearfish, South Dakota, rose from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\) in just 2 minutes. What was the temperature change in Celsius degrees and in kelvins? (b) The temperature in Browning, Montana, was \(44.0^{\circ} \mathrm{Fon}\) January \(23,1916,\) and the next day it plummeted to \(-56.0^{\circ} \mathrm{F} .\) What was the temperature change in Celsius degrees and in kelvins?

\(\bullet\) A 0.500 kg chunk of an unknown metal that has been in boiling water for several minutes is quickly dropped into an insulating Styrofoam TM beaker containing 1.00 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 capacity of the metal? (b) Which is more useful for storing energy from heat, 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 capacity you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain your reasoning.

A Styrofoam" bucket of negligible mass contains 1.75 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice. More ice, from a refrigerator at \(-15.0^{\circ} \mathrm{C},\) is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.778 \(\mathrm{kg} .\) Assuming no heat exchange with the surroundings, what mass of ice was added?

. A thermos for liquid helium. A physicist uses a cylindrical metal can 0.250 \(\mathrm{m}\) high and 0.090 \(\mathrm{m}\) in diameter to store liquid helium at \(4.22 \mathrm{K} ;\) at that temperature the heat of vaporization of helium is \(2.09 \times 10^{4} \mathrm{J} / \mathrm{kg} .\) Completely surrounding the metal can are walls maintained at the temperature of liquid nitrogen, 77.3 \(\mathrm{K}\) , with vacuum between the can and the surrounding walls. How much helium is lost per hour? The emissivity of the metal can is 0.200 . The only heat transfer between the metal can and the surrounding walls is by radiation.

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