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Thermodynamics question: density, linear expansion and temperature

An object has a density of 1250 kg m3 at 10C and a coefficient of linear expansion of α = 2.5 × 10−5 1 /K . What is the object’s density when the temperature is 25C?

I have tried using the equation ΔL=αL° ΔT but this equation has nothing to do with density.

the answer should be: 1248.6

Your help would be greatly appreciated!
 
An object has a density of 1250 kg m3
This is a volume mass density, which should have units of mass / volume. The expansion of the object is not restricted to one dimension.

I have tried using the equation ΔL=αL° ΔT but this equation has nothing to do with density.
This is not true. Any change in dimensions will be accompanied by a change in density. Why is this? Is the object's mass changing? How are density and volume related? Is there an equation that fits better for a change in volume?

Good luck.
 
This is a volume mass density, which should have units of mass / volume. The expansion of the object is not restricted to one dimension.



This is not true. Any change in dimensions will be accompanied by a change in density. Why is this? Is the object's mass changing? How are density and volume related? Is there an equation that fits better for a change in volume?

Good luck.
Ok thanks! So you could use the equation β=3α and then use the volume expansion formula. The equation that relates density to volume is ρ=M/V... but you don;t know what the mass is.
 
The equation that relates density to volume is ρ=M/V... but you don;t know what the mass is.
Did you try expressing the volume expansion formula in terms of ρ instead of V? You should find a result that is independent of the object's mass.
 
[QUO TE="uselesslemma, post: 5061484, member: 542831"]Did you try expressing the volume expansion formula in terms of ρ instead of V? You should find a result that is independent of the object's mass.[/QUOTE]

Ok so you go....
V=3αVΔT
=3αΔTM/ρ
=3(2.5*10-5)(298-283)M/1250
=9.0*10-7M

Am I on the right track?
 
V=3αVΔT
Check the correctness of this equation by comparing to the one you provided for linear expansion. V should not cancel algebraically...otherwise how would you find ρ?
 
Check the correctness of this equation by comparing to the one you provided for linear expansion. V should not cancel algebraically...otherwise how would you find ρ?
So I ended up using the equation V=(1+β(Tf-Ti)) Vi and substituting v=m/ρ which gave me the right answer!

Thanks for your help!
 
So I ended up using the equation V=(1+β(Tf-Ti)) Vi and substituting v=m/ρ which gave me the right answer!

Thanks for your help!
Why did you do "1+" in your formula?
 
Why did you do "1+" in your formula?
I'm not really sure.... it was on a sheet my professor handed out it's suppose to be a parametrization for the volume or something it should have actually have been written as V(T)≈(1+β(Δt))V0
 
19,164
3,782
I'm not really sure.... it was on a sheet my professor handed out it's suppose to be a parametrization for the volume or something it should have actually have been written as V(T)≈(1+β(Δt))V0
This looks correct to me.

Chet
 
Formula: Δρ=3αρΔT
Δρ=3*2.5*10-5*1250*(25-10)
Δρ=1.406 Kgm3
As temperature increases, volume increases. As volume increases, density decreases.
Thus as temperature increases, density decreases.
Therefore the final density at 25°C=initial density - final density
=1250-1.406
=1248.594 Kgm3
 

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