Question

You’ve got an aluminum bar that is half a meter long with a circular cross-sectional area of 2.50 cm2and an initial temperature of 300 K.

(a) One half of the bar is inserted into a thermally insulated vessel of liquid helium at 4.20K. How much of the helium (in liters) boils off by the time the inserted half of the bar cools to 4.20 K? (Assume the upper half of the bar does not cool yet).

(b) If we maintain the upper half of the bar at 300 K, what is the approximate boil-off rate of the liquid helium after the inserted half has reached 4.2 K?

Use the following values for aluminum:

Specific heat = 0.210 cal/g·◦C

Thermal conductivity = 31.0 J/s·cm·K at 4.20 K (ignore its temperature variation)

Density = 2.70 g/cm3

The density of liquid helium is 0.125 g/cm3.

Answer 1

(a)

first, we need to find the mass of bar dipped

m = AL

m = 2700 * 2.50e-4 * 0.5/2

m = 0.16875 Kg

heat lost by aluminium rod

Q = mcT

Q = 0.16875 * 900 J / Kg K * (300 - 4.20)

Q = 44924.625 J ----------- (1)

heat gained by helium

Q = m_He * L_He

where L_He is heat of fusion

Q = 125 * V * 2.09e4 ----------- (2)

equate (1) and (2)

125 * V * 2.09e4 =   44924.625

V = 0.01719 m³

or

V = 17.19 L

-----------------------------------------------------

(b)

energy supplied

P = kA dT / dx

P = 3100 * 2.50e-4 * (300 - 4.20) / 0.25

P = 916.98 W

therefore,

boil off rate will be

rate = P / L_He

rate = 916.98 / 2.09e4 * 125

rate = 0.35 L/s