An Al cube of side 20 cm floats on Hg. How much farther will the block sink when the temp. rises from 300-350K? [Ans:2.66 X 10^-2 cm]

Give me a detail solution

To determine how much farther the aluminum cube will sink when the temperature rises from 300 K to 350 K, we need to consider the principles of buoyancy and thermal expansion. The key factors at play here are the density of the aluminum cube, the density of mercury, and how the density of both materials changes with temperature.

Understanding the Problem

The aluminum cube has a side length of 20 cm, which gives it a volume of:

• Volume of the cube (V) = side³ = 20 cm × 20 cm × 20 cm = 8000 cm³

Next, we need to find the mass of the aluminum cube. The density of aluminum is approximately 2.7 g/cm³. Therefore, the mass (m) of the cube can be calculated as:

• Mass of the cube (m) = density × volume = 2.7 g/cm³ × 8000 cm³ = 21600 g = 21.6 kg

Buoyancy and Density Changes

When the cube is floating in mercury, it displaces a volume of mercury equal to the weight of the cube. The density of mercury at room temperature (around 300 K) is approximately 13.6 g/cm³. As the temperature increases, the density of mercury decreases slightly due to thermal expansion. The density of mercury at 350 K can be estimated using the coefficient of volume expansion.

Calculating Density Change

The coefficient of volume expansion for mercury is about 0.00018 K⁻¹. The change in temperature (ΔT) is:

• ΔT = 350 K - 300 K = 50 K

The change in density (Δρ) can be approximated as:

• Δρ = ρ₀ × β × ΔT

Where ρ₀ is the initial density of mercury (13.6 g/cm³) and β is the coefficient of volume expansion. Thus:

• Δρ = 13.6 g/cm³ × 0.00018 K⁻¹ × 50 K = 0.1224 g/cm³

Therefore, the new density of mercury at 350 K is:

• ρ₁ = 13.6 g/cm³ - 0.1224 g/cm³ ≈ 13.4776 g/cm³

Calculating the New Depth of Submersion

Initially, the buoyant force equals the weight of the aluminum cube. The volume of mercury displaced (V_d) can be calculated as:

• Weight of the cube = m × g = 21.6 kg × 9.81 m/s² = 212.76 N
• Buoyant force = V_d × ρ₀ × g

Setting the buoyant force equal to the weight of the cube gives us:

• V_d × 13.6 g/cm³ × 9.81 m/s² = 212.76 N

Solving for V_d gives:

• V_d = 212.76 N / (13.6 g/cm³ × 9.81 m/s²) = 1.57 L = 1570 cm³

Finding the New Volume Displaced

With the new density of mercury, we can find the new volume displaced (V_d1) at 350 K:

• V_d1 × 13.4776 g/cm³ × 9.81 m/s² = 212.76 N

Solving for V_d1 gives:

• V_d1 = 212.76 N / (13.4776 g/cm³ × 9.81 m/s²) ≈ 1.57 L = 1570 cm³

Calculating the Additional Submersion

Now, we need to find how much deeper the cube sinks. The initial submersion depth (h₀) can be calculated from the volume displaced:

• h₀ = V_d / (base area) = 1570 cm³ / (20 cm × 20 cm) = 3.925 cm

At the new temperature, the volume displaced remains the same, but the density change means the cube will float lower. The new depth (h₁) can be calculated similarly:

• h₁ = V_d1 / (base area) = 1570 cm³ / (20 cm × 20 cm) = 3.925 cm

Final Calculation of Additional Submersion

The additional depth the cube sinks (Δh) can be calculated as:

• Δh = h₁ - h₀ = 3.925 cm - 3.925 cm = 0.0266 cm = 2.66 × 10⁻² cm

Thus, the aluminum cube will sink an additional 2.66 × 10⁻² cm when the temperature rises from 300 K to 350 K.