Question

An aluminum rod 0.500 m in length and with a cross-sectional area of 2.20 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

Answer 1

Answer:

a). V$_{He}$   = 14.6208 litres : b). = 0.29844 Litres/sec

Step-by-step explanation:

a).

Step 1.

Volume when one half of rod is inserted = (50/2)*(2.2) = 55cm³

Step 2.

As we know that,

V$_{He}$ = (ρV$_{c}$IΔT)$_{Al}$ / (ρL$_{v}$)$H_{c}$                     where $H_{c}$ = 290 - 4.2 = 285.8K

                                                        sepecififc heat of aluminum = 0.215 cal/g

                                                                                      density = 2.7 g/cm3

      = (2.7)*(55)*(0.215)*(285.8) / (0.125)*(20900)*(1/4186)

V$_{He}$   = 14.6208 litres

b).

Step 1.

As we know that

  P = kA(dT/dx) = 31*2.2*[(290-4.2)/25]

                         = 779.6624 W

Step 2.

Now, approximate boil rate of helium is:

P/ρL$_{v}$= [(779.6624)*(10^3g/kg)] / [(0.125)*(20900)]

         = 0.29844 Litres/sec