A 1-liter aluminum container at 20o

C is filled with 0.975 liters of mercury at
20o
C. If both the container and the mercury are heated, at what final temperature
will the mercury completely fill the container? (Hint: this is a volume expansion
problem)
αAl = 24 × 10-6
(Cº
)
-1
; βHg = 1.82×10-4
(°C)-1

To solve this problem, we can use the concept of thermal expansion and the formula for volume expansion.

The volume expansion formula is given by:

ΔV = V₀ * β * ΔT

Where:
ΔV is the change in volume
V₀ is the initial volume
β is the coefficient of volume expansion
ΔT is the change in temperature

In this case, we want the mercury to completely fill the container, so the change in volume, ΔV, will be equal to the initial volume, V₀.

Given:
Initial volume, V₀ = 0.975 liters
Coefficient of volume expansion for aluminum, αAl = 24 × 10^-6 (Cº)^-1
Coefficient of volume expansion for mercury, βHg = 1.82 × 10^-4 (°C)^-1

Let's assume the final temperature is T.

For the aluminum container, the change in volume, ΔV_al, is given by:
ΔV_al = V_al - V₀
Since we want the mercury to completely fill the container, ΔV_al will be equal to zero.

For the mercury, the change in volume, ΔV_Hg, is given by:
ΔV_Hg = V_Hg - V₀

Now, let's set up the equations:

For aluminum:
0 = V_al - V₀
V_al = V₀

For mercury:
ΔV_Hg = V_Hg - V₀
Using the volume expansion formula:
ΔV_Hg = V₀ * βHg * (T - 20)

Substituting V_al = V₀ into the equation for mercury:
ΔV_Hg = V_al * βHg * (T - 20)
0.975 = 0.975 * 1.82 × 10^-4 * (T - 20)

Now, solve for T:

0.975 = 0.975 * 1.82 × 10^-4 * T - 0.975 * 1.82 × 10^-4 * 20
1 = 1.82 × 10^-4 * T - 1.82 × 10^-4 * 20
1 + 1.82 × 10^-2 = 1.82 × 10^-4 * T
(1 + 1.82 × 10^-2) / 1.82 × 10^-4 = T
(5.4945 × 10^4) / (1.82 × 10^-4) = T

Calculating the value:
T ≈ 301.89

Therefore, the final temperature at which the mercury completely fills the container is approximately 301.89°C.