Answer:
[tex]\dot m = 3.33 kg/hr[/tex]
Explanation:
Given data:
[tex]T_{He} = 4.2 K[/tex]
[tex]T_s = 58.4 K[/tex]
Radius of contaner = 0.589 m
emissivity is = 1
surface area =[tex] 4\pi r^2 = 4\pi *0.589^2 = 4.35 m^2[/tex]
Net Radiation between heilum and shell is given as
[tex]\dot Q = \epsilon \sigma A(T_s^4 _T_{he}^4)[/tex]
[tex] =1\times 5.67\times 10^{-8} \times 4.35 (58.4^4 -4.2^4)[/tex]
[tex]\dot Q = 2.88 J/s [/tex]
for 6.76 hr = 70,087.68 J/hr
rate of energy removed can be calculated by using the following relation
[tex]\dot Q = \dot m L[/tex]
[tex]70,087.68 = \dot m \times 2.1\times 10^4[/tex]
solving for \dot m we get
[tex]\dot m = 3.33 kg/hr[/tex]
