A hollow aluminum cylinder 22.5 cm deep has an internal capacity of 2.000 L at 16.0°C. It is completely filled with turpentine at 16.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 70.0°C. (The average linear expansion coefficient for aluminum is 2.4 10-5°C−1, and the average volume expansion coefficient for turpentine is 9.0 10-4°C−1.)

(a) How much turpentine overflows?
(b) What is the volume of turpentine remaining in the cylinder at 70.0°C? (Give you answer to four significant figures.)
(c) If the combination with this amount of turpentine is then cooled back to 16.0°C, how far below the cylinder's rim does the turpentine's surface recede?

To answer these questions, we need to consider the expansion of both the aluminum cylinder and the turpentine.

(a) To calculate the amount of turpentine that overflows, we need to find the increase in volume of the aluminum cylinder and the increase in volume of the turpentine.

The increase in volume of the aluminum cylinder can be calculated using the linear expansion formula:
ΔV = V₀ * α * ΔT,

where ΔV is the change in volume, V₀ is the initial volume, α is the linear expansion coefficient, and ΔT is the change in temperature.

Given:
Initial volume of the aluminum cylinder, V₀ = 2.000 L = 2000 cm³,
Linear expansion coefficient of aluminum, α_aluminum = 2.4 * 10^(-5) °C^(-1),
Change in temperature, ΔT = 70.0 °C - 16.0 °C = 54.0 °C.

Substituting these values into the formula, we get:
ΔV_aluminum = V₀ * α_aluminum * ΔT
= 2000 cm³ * (2.4 * 10^(-5) °C^(-1)) * 54.0 °C.

Now, let's calculate the increase in volume of turpentine.

The increase in volume of the turpentine can be calculated using the volume expansion formula:
ΔV = V₀ * β * ΔT,

where ΔV is the change in volume, V₀ is the initial volume, β is the volume expansion coefficient, and ΔT is the change in temperature.

Given:
Initial volume of the turpentine, V₀ = 2000 cm³, (since it fills the aluminum cylinder completely)
Volume expansion coefficient of turpentine, β_turpentine = 9.0 * 10^(-4) °C^(-1),
Change in temperature, ΔT = 70.0 °C - 16.0 °C = 54.0 °C.

Substituting these values into the formula, we get:
ΔV_turpentine = V₀ * β_turpentine * ΔT
= 2000 cm³ * (9.0 * 10^(-4) °C^(-1)) * 54.0 °C.

To calculate the amount of turpentine that overflows, we subtract the increase in volume of the aluminum cylinder from the increase in volume of the turpentine:
Amount of turpentine that overflows = ΔV_turpentine - ΔV_aluminum.

Now, we can substitute the calculated values and solve for the amount of turpentine that overflows.

(b) To calculate the volume of turpentine remaining in the cylinder at 70.0 °C, we subtract the volume expansion of the turpentine from the initial volume of the turpentine.

Initial volume of the turpentine = 2000 cm³ (given),
Change in volume of the turpentine = ΔV_turpentine (calculated in part a).

Volume of turpentine remaining = Initial volume of turpentine - Change in volume of turpentine.

(c) To calculate how far below the cylinder's rim the turpentine's surface recedes when cooled back to 16.0 °C, we need to consider the decrease in volume of the turpentine and the decrease in volume of the aluminum cylinder.

The decrease in volume of the turpentine can be calculated using the volume expansion formula:
ΔV = V₀ * β * ΔT,

where ΔV is the change in volume, V₀ is the initial volume, β is the volume expansion coefficient, and ΔT is the change in temperature.

Given:
Initial volume of the turpentine = Volume of turpentine remaining (calculated in part b),
Volume expansion coefficient of turpentine, β_turpentine = 9.0 * 10^(-4) °C^(-1),
Change in temperature, ΔT = 70.0 °C - 16.0 °C = 54.0 °C.

To calculate the decrease in volume of the aluminum cylinder, we use the same linear expansion formula as in part a.

The decrease in volume of the aluminum cylinder = ΔV_aluminum (calculated in part a).

To calculate how far below the cylinder's rim the turpentine's surface recedes, we subtract the decrease in volume of the turpentine from the decrease in volume of the aluminum cylinder:
Decrease in turpentine's surface = ΔV_aluminum - ΔV_turpentine.

Now, let's calculate the values and solve each part one by one.

To solve this problem, we need to consider the expansion of both the aluminum cylinder and the turpentine as they are heated.

(a) To find the amount of turpentine that overflows, we first need to calculate the change in volume of the aluminum cylinder as it is heated. We can use the formula for linear expansion:

ΔV = V0 * α * ΔT

Where:
ΔV is the change in volume of the aluminum cylinder,
V0 is the initial volume of the aluminum cylinder,
α is the linear expansion coefficient of aluminum, and
ΔT is the change in temperature.

The change in temperature ΔT is the final temperature minus the initial temperature:

ΔT = 70.0°C - 16.0°C = 54.0°C

The linear expansion coefficient for aluminum is given as 2.4 * 10^-5 °C^(-1).

The initial volume of the aluminum cylinder is the same as its internal capacity, given as 2.000 L. To convert this to cubic centimeters (cm^3), we multiply by 1000:

V0 = 2.000 L * 1000 cm^3/L = 2000 cm^3

Now, we can calculate ΔV:

ΔV = 2000 cm^3 * 2.4 * 10^-5 °C^(-1) * 54.0°C
= 259.2 cm^3

Since the cylinder is completely filled with turpentine, this amount of turpentine will overflow.

(b) To find the volume of turpentine remaining in the cylinder at 70.0°C, we subtract the amount of turpentine that overflows from the initial volume of the cylinder:

V_remaining = V0 - ΔV
= 2000 cm^3 - 259.2 cm^3
= 1740.8 cm^3

The volume of turpentine remaining in the cylinder at 70.0°C is approximately 1740.8 cm^3.

(c) To determine how far below the cylinder's rim the turpentine's surface recedes when cooled back to 16.0°C, we need to calculate the change in volume of turpentine due to its expansion/contraction.

The volume expansion coefficient for turpentine is given as 9.0 * 10^-4 °C^(-1).

We can use the same formula for volume expansion as before:

ΔV_turpentine = V_turpentine0 * β * ΔT

Where:
ΔV_turpentine is the change in volume of turpentine,
V_turpentine0 is the initial volume of turpentine,
β is the volume expansion coefficient of turpentine, and
ΔT is the change in temperature.

The change in temperature ΔT is again 70.0°C - 16.0°C = 54.0°C.

The initial volume of turpentine is the same as the remaining volume from part (b), 1740.8 cm^3.

ΔV_turpentine = 1740.8 cm^3 * 9.0 * 10^-4 °C^(-1) * 54.0°C
≈ 840.31 cm^3

Therefore, the turpentine's surface will recede approximately 840.31 cm^3 below the cylinder's rim when cooled back to 16.0°C.