Answer:
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
