An aluminum rod of length 3.3 m and crosssectional area 3.8 cm2 is inserted vertically into a thermally insulated vessel containing liquid helium at 4.2 K. The rod is initialy at 274 K. If half of the rod is inserted into the helium, how many liters of helium boil off in the very short time during which the inserted half cools to 4.2 K? The density of liquid helium at 4.2 K is 122 kg/m3 and its latent heat is 20900 J/kg. The specific heat of aluminum is 900 J/kg · ◦ C and its density is 2700 kg/m3 . Answer in units of L.

V_vap = 161.2 L

Explanation:

The total mass of the aluminum rod is given as;

m = ρ∙V = ρ∙L∙A

Where;

ρ is density = 2700 kg/m³

L is length = 3.3m

A is cross sectional area = 3.8 cm² = 3.8 x 10⁻⁴ m²

Thus;

m = 2700kg/m³•3.3m•3.8 × 10⁻⁴m²

= 3.3858kg

By cooling down the submerged half of the aluminum rod releases an heat amount of

Q = (1/2)∙m∙cp∙∆T

Where;

cp is specific heat of aluminum aluminum = 900 J/kg

∆T is change in temperature = 274 - 4.2 = 269.8 K

Thus;

Q = (1/2)•3.3858•900•(269.8)

= 411069.978 J

The liquid absorbs this heat and vaporizes partially, such that the heat equals vaporized mass times latent heat of vaporization:

Q = m_vap•∆h_vap

Making m_vap the subject;

m_vap∙ = Q/∆h_vap

Where ∆h_vap is latent heat of vaporization given as 20900J/kg

Thus,

m_vap∙ = 411069.978/20900

= 19.668 kg

Let's divide this mass by the density of liquid helium and we get the liquid volume which has vaporized:

V_vap∙= m_vap/ρ

V_vap∙ = 19.668/122

V_vap∙ = 0.1612 m³

Converting to litres;

V_vap = 0.1612 x 1000

V_vap = 161.2 L

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