An aluminum rod 0.500 m in length and with a cross-sectional area of 2.40 cm 2 i

Question

An aluminum rod 0.500 m in length and with a cross-sectional area of 2.40 cm2 is inserted into a thermally insulated vessel containing liquid helium at 4.20 K. The rod is initially at 290 K. (Aluminum has thermal conductivity of 3 100 W/m · K at 4.20 K; ignore its temperature variation. Aluminum has a specific heat of 0.215 cal/g·°C and density of 2.70 g/cm3. The density of liquid helium is 0.125 g/cm3.)

(a) 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.
liters

(b) If the circular surface of the upper end of the rod is maintained at 290 K, what is the approximate boil-off rate of liquid helium after the lower half has reached 4.20 K?
liters/s

Explanation / Answer

a) Specific heat of alluminium =0.215 cal/g·°C=900 J/kg.oC

The latent heat of helium 20900 J/kg

The total mass of the aluminum rod is

m = V = LA
= 2700kg/m³ 0.500m 2.40×10-4m²
= 0.324 kg

By cooling down the submerged half of the aluminum rod releases an heat amount of
Q = (1/2)m.CpT
= (1/2) 0.324kg 900J/kg (290K - 4.2K)
= 41669.64J

The liquid absorbs this heat and vaporizes partially, such that the heat equals vaporized mass times latent heat of vaporization:
Q = m_vaph_vap
=> m_vap= Q/h_vap

= 41669.64J / 20900(J/kg)

= 1.994kg

divide this mass by the density of liquid helium and you get the liquid volume which has vaporized:
V_vap= m_vap/
= 1.994kg / 125kg/m³
= 0.0159m³
= 15.952L

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