An aluminum can is filled to the brim with a liquid. The can and the liquid are
heated so their temperatures change by the same amount. The can’s initial volume
at 15 ºC is 4.5×10-4 m3
. The coefficient of volume expansion for aluminum is
6.9×10-5
ºC-1
. When the can and the liquid are heated to 75 ºC, 2.9×10-6 m3
of the
liquid spills over.
a. Calculate the coefficient of volume expansion of the liquid. [10]
b. Find the coefficient of linear expansion for the aluminum can. Include
statements of any assumptions made. [4]
a. To calculate the coefficient of volume expansion of the liquid, we can use the equation:
ΔV = V0 * β * ΔT
Where:
ΔV is the change in volume of the liquid
V0 is the initial volume of the liquid
β is the coefficient of volume expansion of the liquid
ΔT is the change in temperature
ΔV = 2.9×10-6 m3 (given)
V0 = 4.5×10-4 m3 (given)
ΔT = 75 ºC - 15 ºC = 60 ºC
Substituting the known values into the equation, we can solve for β:
2.9×10-6 m3 = 4.5×10-4 m3 * β * 60 ºC
β = 2.9×10-6 m3 / (4.5×10-4 m3 * 60 ºC)
β ≈ 3.02×10-9 ºC-1
Therefore, the coefficient of volume expansion of the liquid is approximately 3.02×10-9 ºC-1.
b. To find the coefficient of linear expansion for the aluminum can, we need to make the assumption that the can's change in volume is negligible compared to the change in volume of the liquid. This assumption is reasonable since the liquid spilling over is a relatively small amount compared to the initial volume of the can.
Using the formula:
ΔV = V0 * α * ΔT
Where:
ΔV is the change in volume of the can
V0 is the initial volume of the can
α is the coefficient of linear expansion of the can
ΔT is the change in temperature
We can rearrange the equation to solve for α:
α = ΔV / (V0 * ΔT)
Substituting the known values:
ΔV = 0 (negligible change in volume of the can)
V0 = 4.5×10-4 m3 (given)
ΔT = 75 ºC - 15 ºC = 60 ºC
We find:
α = 0 / (4.5×10-4 m3 * 60 ºC)
α = 0 m-1 ºC-1
The coefficient of linear expansion for the aluminum can is approximately 0 m-1 ºC-1.
a. To calculate the coefficient of volume expansion of the liquid, we can use the equation:
ΔV = βV0ΔT
Where:
ΔV is the change in volume of the liquid,
β is the coefficient of volume expansion of the liquid,
V0 is the initial volume of the liquid,
and ΔT is the change in temperature.
Given:
- Initial volume of the can (V0) = 4.5×10^(-4) m^3
- Change in temperature (ΔT) = 75 ºC - 15 ºC = 60 ºC
- Change in volume of liquid (ΔV) = 2.9×10^(-6) m^3
Using the equation, we can rearrange to solve for β:
β = ΔV / (V0 * ΔT)
Plugging in the values:
β = (2.9×10^(-6) m^3) / (4.5×10^(-4) m^3 * 60 ºC)
Simplifying the expression:
β = 2.9×10^(-6) / (4.5×10^(-4) * 60)
β ≈ 1.018×10^(-4) ºC^(-1)
Therefore, the coefficient of volume expansion of the liquid is approximately 1.018×10^(-4) ºC^(-1).
b. To calculate the coefficient of linear expansion of the aluminum can (α), we can use the equation:
ΔL = αL0ΔT
Where:
ΔL is the change in length of the aluminum can (assumed to be linear),
α is the coefficient of linear expansion of aluminum,
L0 is the initial length of the aluminum can,
and ΔT is the change in temperature.
Assuming that the can is a cylinder, we can approximate the change in length as the change in volume divided by the initial cross-sectional area:
ΔL = ΔV / A0
Where:
A0 is the initial cross-sectional area of the can.
Using the equation for the coefficient of linear expansion, we can now solve for α:
α = ΔL / (L0 * ΔT)
Plugging in the values:
α = (ΔV / A0) / (L0 * ΔT)
Since the can is filled to the brim, the volume change is equal to the change in volume of the liquid (2.9×10^(-6) m^3), so:
α = (2.9×10^(-6) m^3 / A0) / (L0 * ΔT)
However, we do not have enough information to directly calculate α since we do not have values for A0 and L0. Additional assumptions or measurements would be needed to determine the coefficient of linear expansion for the aluminum can.
Assuming that the can is a cylindrical shape, we can make the further assumption that the shape of the cross-section remains constant during the heating process, implying that the change in length is directly proportional to the change in volume and can be approximated with a linear expansion coefficient equivalent to that of aluminum (2.3×10^(-5) ºC^(-1)).