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.
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]
1 answer
1 answer