An aluminum rod 0.500 m in length and with a cross-sectional area of 2.50 cm^2 is inserted into a thermally insulated vessel containing liquid helium at 4.20 K (B.P. for helium). The rod is initially at 300 K. (𝑐_𝐴𝑙=900 𝐽/𝑘𝑔°𝐶, 𝐿_(𝑣,𝐻𝑒)=2.09×10^4 𝐽/𝑘𝑔)

Question

An aluminum rod 0.500 m in length and with a cross-sectional area of 2.50 cm^2 is inserted into a thermally insulated vessel containing liquid helium at 4.20 K (B.P. for helium). The rod is initially at 300 K. (𝑐_𝐴𝑙=900 𝐽/𝑘𝑔°𝐶, 𝐿_(𝑣,𝐻𝑒)=2.09×10^4 𝐽/𝑘𝑔)

If one half of the rod is inserted into the helium, how many liters of helium boil off by the time the inserted half cools to 4.20 K? Assume the upper half does not yet cool.

Answer 1

To solve this problem, we need to calculate the heat transfer from the aluminum rod to the liquid helium until the rod cools to 4.20 K.

Step 1: Calculate the mass of the aluminum rod.
Given: length of the rod (L) = 0.500 m, cross-sectional area (A) = 2.50 cm^2 = 2.50 × 10^-4 m^2
Density of aluminum (ρ) = 2700 kg/m^3 (approximate value)
Mass of the aluminum rod (m) = ρ × V, where V is the volume of the rod.
V = A × L/2 (assuming half of the rod is inserted)
m = 2700 kg/m^3 × 2.50 × 10^-4 m^2 × 0.500 m/2

Step 2: Calculate the heat transfer from the aluminum rod to the liquid helium.
Heat transfer (Q) = m × specific heat capacity of aluminum (c_Al) × change in temperature (ΔT)
ΔT = 300 K - 4.20 K

Step 3: Calculate the amount of helium that boils off.
The heat required to boil off a certain amount of helium can be calculated using the latent heat of vaporization of helium (L_v,He).
Q = m_He × L_v,He
where m_He is the mass of helium vaporized and L_v,He is the latent heat of vaporization of helium.

Step 4: Calculate the volume of helium that boils off.
Given: specific latent heat of vaporization of helium (L_v,He) = 2.09 × 10^4 J/kg
Density of helium (ρ_He) = 125.98 kg/m^3 (approximate value)
Volume of helium (V_He) = m_He / ρ_He

Let's perform the calculations:

Step 1:
m = 2700 kg/m^3 × 2.50 × 10^-4 m^2 × 0.500 m/2
m ≈ 0.16875 kg

Step 2:
ΔT = 300 K - 4.20 K
ΔT ≈ 295.80 K

Q = m × c_Al × ΔT
Q = 0.16875 kg × 900 J/kg°C × 295.80 K

Step 3:
Q = m_He × L_v,He
m_He ≈ Q / L_v,He

Step 4:
V_He = m_He / ρ_He

Now, substitute the values and calculate:

m_He ≈ (0.16875 kg × 900 J/kg°C × 295.80 K) / (2.09 × 10^4 J/kg)
V_He ≈ m_He / ρ_He

Please note that the equations and constants used in this answer are approximate and may differ slightly in actual practice.