Does density change with temperature and how would you calculate it?

  • Thread starter Thread starter nemzy
  • Start date Start date
  • Tags Tags
    Density
AI Thread Summary
Density changes with temperature due to the expansion or contraction of substances, affecting their volume. The average coefficient of volume expansion can be used to calculate the change in volume with temperature changes. The formula ΔV = δ * V * ΔT helps determine the new volume based on the original volume and temperature change. To find the new density at a different temperature, the mass remains constant, allowing the calculation of density using the formula density = mass / volume. By applying these principles, one can determine the density of gasoline at various temperatures.
nemzy
Messages
124
Reaction score
0
lets say for example the density for gasoline is 730 kg/m^3 at 0 degree Celsisus

Will the density be different for at a diff temperature, or will it always remain constant?

If it is different, how would you calculate it? thanks
 
Physics news on Phys.org
Density is mass per unit volume (well, mass density anyways)

So heating or cooling would change the density beucase the volume that the substance would change. There are a number of different ways to find the change in volume of the substance...depending on what form it is in, etc.
 
well for my homework question, it gives the avg coefficient of volume expansion. and from there on i have no idea how to calculate the density of gasoline at a diff temp.
 
use the following equation:

\Delta V = \delta*V*\Delta T

Where \Delta V is the change in volume, \delta is the coefficient of volume expansion, V is the current volume, and \Delta T is the change in temp

Edit: Changed symbols to Latex
 
Last edited:
i know, but once u find the change in volume, how can u relate it to finding density at a certain temp?

for example, let's say density of gasoline is 730 kg/m^3 at 0 degrees

what will the density be at 20 degrees?

a gallon of gasoline occupies .00380 m^3 and the avg coefficient of volume expansion is 9.6e-4

So plugging it into the equation, i got a change of volume of 7.296e-5, with final volume being .00387296...now from here how would i be able to find the new density at the new temp?
 
well, remember, mass density = mass / volume

you know the new volume...and you know the mass (hint: use the original density and the original volume to find the mass)
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top