How to Calculate Thermodynamic Equilibrium using Basic Principles

In summary, the rail temperature rises by 30 degrees Celsius, but the middle joint between the two parts of the rail rises upwards by much more due to the temperature change. The thermal expansion coefficient of steel is 1.2×10-5 K-1, so the middle joint would rise by 0.108 light years due to the temperature change.
  • #1
xiaoyaozizai
4
0
Homework Statement
A 30 m long steel rail is firmly secured to the ground at both ends . The rail temperature rises by 30 ° C in hot sunshine. Assume that the rail consists of two equally long parts that are articulated in the middle. How much does the middle joint rise upwards due to temperature change ? The thermal expansion coefficient of the steel is 1,2 × 10-5 K-1 .
Relevant Equations
I used the

30× (1.2×10-5) ×30

= 0.108

That's not the right answer. Did I use the wrong formula
30×(1.2×10-5)×30

= 0.108

Help
 
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  • #2
You have calculated the amount by which the total length increases. By how much it rises is the question.
That depends on the shape. If it were a single length of rail, that is moderately advanced to calculate, but if I inderstand the question, because of the articulation the shape is two straight lines ∧ which is much easier.
 
  • #3
Where did you get that crazy formula from?
 
  • #4
Chestermiller said:
Where did you get that crazy formula from?
That gives some additional error.
 
  • #5
xiaoyaozizai said:
Homework Statement:: A 30 m long steel rail is firmly secured to the ground at both ends . The rail temperature rises by 30 ° C in hot sunshine. Assume that the rail consists of two equally long parts that are articulated in the middle. How much does the middle joint rise upwards due to temperature change ? The thermal expansion coefficient of the steel is 1,2 × 10-5 K-1 .
Relevant Equations:: I used the

30× (1.2×10-5) ×30

= 0.108

That's not the right answer. Did I use the wrong formula
0.108 light years? Yeah, that's pretty unreasonable.

In the formula ##\Delta L = L_0 \alpha \Delta T##, what does ##\Delta L## represent? Is it what's being asked for in the problem?

Physics isn't about plugging numbers into some equation. It's about analyzing a situation and applying relatively few concepts to get to the desired result. You need to work on the analysis part.
 
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Likes jbriggs444 and Lnewqban
  • #6
vela said:
0.108 光年?是的,这很不合理。

在公式##\Delta L = L_0 \alpha \Delta T##中,##\Delta L##代表什么?这是问题中所要求的吗?

物理学不是将数字代入某个方程式。它是关于分析情况并应用相对较少的概念来达到预期的结果。您需要处理分析部分。
ok thank you
 
  • #7
Chestermiller said:
Where did you get that crazy formula from?
30×(1.2×10-5)×30 appears to be an attempt to write ##30 \text{ meters} \times \frac{1.2 \times 10^{-5}}{\text{K}} \times 30 \text{ K}## to obtain, supposedly, ##0.108 \text{ meters}##

However, taking 900 (close to 1000) and multiplying it by 1.2/100000 (close to 1/100000) does not yield something close to 1/10. Mathematical literacy allows one to sometimes notice when calculators return the wrong answer.

There is an order of magnitude error even before we get to the point of solving the correct problem. That part should be another couple of orders of magnitude to deal with.
 
  • #8
jbriggs444 said:
30×(1.2×10-5)×30 似乎是尝试写 ##30 \text{meter} \times \frac{1.2 \times 10^{-5}}{\text{K}} \times 30 \text{ K}## 得到,据说,##0.108 \text{ 米}##

但是,将 900(接近 1000)乘以 1.2/100000(接近 1/100000)不会得到接近 1/10 的结果。当计算器返回错误答案时,数学素养有时会让人注意到。

甚至在我们到达解决正确问题的地步之前,就有一个数量级的误差。这部分应该是另外几个数量级的处理。
ok i know thank you for explaining
 

Related to How to Calculate Thermodynamic Equilibrium using Basic Principles

1. What is thermodynamics?

Thermodynamics is a branch of physics that deals with the study of heat and its relation to energy and work. It also involves the behavior of materials and systems at different temperatures and how they change in response to heat and energy transfer.

2. Why is thermodynamics considered basic?

Thermodynamics is considered basic because it is a fundamental principle that governs the behavior of all physical systems. It is essential in understanding the laws of energy conservation, entropy, and the transfer of heat and energy in various processes.

3. What are the laws of thermodynamics?

The laws of thermodynamics are a set of fundamental principles that govern the behavior of energy and heat in physical systems. They include the first law, which states that energy cannot be created or destroyed, only transferred or converted, and the second law, which states that the total entropy of a closed system will always increase over time.

4. How is thermodynamics used in everyday life?

Thermodynamics has numerous applications in everyday life, from heating and cooling systems to cooking and power generation. It is also used in the design of engines, refrigerators, and other machines that involve energy transfer and conversion.

5. What are the different branches of thermodynamics?

There are three main branches of thermodynamics: classical thermodynamics, which deals with macroscopic systems and their properties; statistical thermodynamics, which explains the behavior of systems at the microscopic level; and quantum thermodynamics, which combines the principles of quantum mechanics with thermodynamics to describe the behavior of small systems.

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