Show that ∆ρ = -ρβ∆t
We know that the change in density (∆ρ) of a material depends on its coefficient of volumetric thermal expansion (β) and the change in temperature (∆t).
The relationship between these variables can be expressed as:
∆V = Vβ∆t
where ∆V is the change in volume of the material, V is its original volume, β is the coefficient of volumetric thermal expansion, and ∆t is the change in temperature.
We also know that density (ρ) is the mass (m) of a material divided by its volume:
ρ = m/V
Taking the derivative of this equation with respect to time gives:
∆ρ/∆t = (dm/dt)/V - m/(V^2) * (dV/dt)
Since we are assuming that the mass of the material is constant, dm/dt = 0, and the first term on the right-hand side of the equation is zero.
Putting in the expression for ∆V and simplifying, we get:
∆ρ/∆t = -ρβ∆t
Therefore,
∆ρ = -ρβ∆t
which shows that the change in density (∆ρ) is proportional to the negative product of the original density (ρ), coefficient of volumetric thermal expansion (β), and change in temperature (∆t).
To show that ∆ρ = -ρβ∆t, where ∆ρ is the change in density, ρ is the initial density, β is the volume coefficient of thermal expansion, and ∆t is the change in temperature, we can use the formula for volume expansion and the definition of density.
The formula for volume expansion is:
∆V = Vβ∆t
where ∆V is the change in volume, V is the initial volume, and β is the volume coefficient of thermal expansion.
Density is defined as the mass divided by the volume:
ρ = m/V
where ρ is the density, m is the mass, and V is the volume.
We can rewrite the formula for density as:
m = ρV
Now, let's consider the change in density. The change in density (∆ρ) can be calculated by taking the final density (ρ + ∆ρ) and subtracting the initial density (ρ):
∆ρ = (ρ + ∆ρ) - ρ
Simplifying this expression, we get:
∆ρ = ∆ρ
Next, we will relate the change in density to the change in volume (∆V) and the change in temperature (∆t).
Using the formula for volume expansion (∆V = Vβ∆t), we can express the change in volume (∆V) as a function of the initial volume (V), the volume coefficient of thermal expansion (β), and the change in temperature (∆t):
∆V = Vβ∆t
Now, we will express the initial volume (V) in terms of the initial density (ρ) and the mass (m) using the definition of density (ρ = m/V):
V = m/ρ
Substituting this expression for V in the formula for volume expansion, we get:
∆V = (m/ρ)β∆t
Next, we will express the change in mass (∆m) as a function of the initial mass (m) and the change in density (∆ρ):
∆m = (m + ∆m) - m
Simplifying this expression, we get:
∆m = ∆m
Now, let's express the change in mass (∆m) in terms of the initial mass (m) and the change in density (∆ρ) using the definition of density (ρ = m/V):
∆m = (∆ρ)V
Substituting the expression for V in terms of ρ, we get:
∆m = (∆ρ)(m/ρ)
Now, let's express the change in volume (∆V) in terms of the change in mass (∆m) using the definition of density (ρ = m/V):
∆V = (∆m)/ρ
Substituting the previously derived expression for ∆m, we get:
∆V = (∆ρ)(m/ρ)/ρ
Now, let's express the change in density (∆ρ) in terms of the change in volume (∆V) and the initial density (ρ):
∆ρ = (∆V)(ρ²/m)
By substituting ∆V = Vβ∆t and ρ = m/V into the above expression, we get:
∆ρ = (Vβ∆t)(m²/V²m) = β∆t(m/V) = β∆tρ
Finally, we have:
∆ρ = β∆tρ
Which is the desired result.
Therefore, we have shown that ∆ρ = -ρβ∆t.