Change in density due to thermal expansion

In summary, the conversation discusses finding a mathematical proof for the change in density of an object due to thermal expansion. The equations V=(1+βΔT)Vi, L=(1+αΔT)Li, and V=m/ρ are mentioned as relevant. The solution involves using a cube and simplifying equations to ultimately yield Δρ/ρ = -3(ΔL/L), with a negative sign being implied.
  • #1
empty_set
2
0

Homework Statement


I want to find the mathematical proof to show that the density of an object changes with thermal expansion. My professor showed this in class and it was horribly wrong because he let a few things out. The book I'm reading showed it in 4 steps and left out a lot of the crucial parts. I've been working on this for a few days.

How do I go from, Δρ=m/ΔV, to Δρ/ρ= -3(ΔL/L) ?
I think these are the relevant equations:
V=(1+βΔT)Vi
L=(1+αΔT)Li
V=m/ρ

The Attempt at a Solution


This is how I got it but I don't think it's correct: m.imgur.com/a/ZOfCp
 
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  • #2
My friend, don't struggle so much for it.

You know that ρ=m/V.
Let ρ°=m/V°.
Then V°= V( 1+γΔT) where γ=3α.
Put that into the equation above, divide by ρ, subtract 1. You have your answer.
 
  • #3
empty_set said:

Homework Statement


I want to find the mathematical proof to show that the density of an object changes with thermal expansion. My professor showed this in class and it was horribly wrong because he let a few things out. The book I'm reading showed it in 4 steps and left out a lot of the crucial parts. I've been working on this for a few days.

How do I go from, Δρ=m/ΔV, to Δρ/ρ= -3(ΔL/L) ?
I think these are the relevant equations:
V=(1+βΔT)Vi
L=(1+αΔT)Li
V=m/ρ

The Attempt at a Solution


This is how I got it but I don't think it's correct: m.imgur.com/a/ZOfCp
It isn't clear what you did. I suggest doing the analysis on a cube. $$V_0=L_0^3$$
$$V=L^3=[L_0(1+\alpha \Delta T)]^3$$
and neglect non-linear terms in ##\Delta T##.
 
  • #4
For both of those methods I got Δρ/ρ = 3(ΔL/L). Would the negative be implied or is my algebra wrong?
 
  • #5
empty_set said:
For both of those methods I got Δρ/ρ = 3(ΔL/L). Would the negative be implied or is my algebra wrong?
There should be a minus sign in this equation.
 
  • Like
Likes CrazyNinja
  • #6
Your algebra is wrong. My method WILL give a minus sign inherently. Check again. You probably got confused with the ρ and ρ°.
 

What is thermal expansion?

Thermal expansion is the tendency of matter to change in volume or density in response to a change in temperature. When heated, most materials expand and when cooled, they contract.

How does thermal expansion affect density?

As a material is heated, the atoms and molecules within it vibrate faster and take up more space, causing the material to expand. This increase in volume leads to a decrease in density, as the same amount of matter is spread out over a larger space.

What is the coefficient of thermal expansion?

The coefficient of thermal expansion (CTE) is a measure of how much a material's volume or density changes in response to a change in temperature. It is typically expressed in terms of a fractional change in length or volume per unit temperature change.

How is the change in density due to thermal expansion calculated?

The change in density due to thermal expansion can be calculated using the formula Δρ = ρ0 x α x ΔT, where Δρ is the change in density, ρ0 is the initial density, α is the coefficient of thermal expansion, and ΔT is the change in temperature.

What are some real-world examples of thermal expansion affecting density?

Some common examples include the expansion of metals in bridges and buildings due to changes in temperature, the expansion of gases inside a hot air balloon causing it to rise, and the expansion of water when frozen, leading to icebergs having less density than liquid water.

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