Thermodynamics question: density, linear expansion and temperature

[Patricia Reid]

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

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!

 

[uselesslemma]

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.

 

[Patricia Reid]

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.

 

[uselesslemma]

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.

 

[Patricia Reid]

[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?

 

[uselesslemma]

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?

 

[Patricia Reid]

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(T_f-T_i)) V_i and substituting v=m/ρ which gave me the right answer!

Thanks for your help!

 

[forceanger]

So I ended up using the equation V=(1+β(T_f-T_i)) V_i and substituting v=m/ρ which gave me the right answer!

Thanks for your help!

Why did you do "1+" in your formula?

 

[Patricia Reid]

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

 

[Chestermiller]

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

 

[asmitha]

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

 

[mohdabbag]

ρ2=ρ1/1+βΔT
use this formula it should work.