Question

Liquid helium is stored at its boiling-point temperature of4.2

K in a spherical container (r = 0.589 m). Thecontainer is
a perfect blackbody radiator. The container issurrounded by a
spherical shield whose temperature is 58.4 K. Avacuum exists in the
space between the container and shield. Thelatent heat of
vaporization for helium is 2.1 x 104J/kg. What mass of
liquid helium boils away through the ventingvalve in 6.76
hours?

Answer 1

`\dot m = 3.33 kg/hr`

Step-by-step explanation:

Given data:

`T_(He) = 4.2 K`

`T_s = 58.4 K`

Radius of contaner = 0.589 m

emissivity is = 1

surface area =
`4\pi r^2 = 4\pi *0.589^2 = 4.35 m^2`

Net Radiation between heilum and shell is given as

`\dot Q = \epsilon \sigma A(T_s^4 _T_(he)^4)`

`=1* 5.67* 10^(-8) * 4.35 (58.4^4 -4.2^4)`

`\dot Q = 2.88 J/s`

for 6.76 hr = 70,087.68 J/hr

rate of energy removed can be calculated by using the following relation

`\dot Q = \dot m L`

`70,087.68 = \dot m * 2.1* 10^4`

solving for \dot m we get

`\dot m = 3.33 kg/hr`