Question

Liquid helium is stored at its boiling point $\left(\right.4.2\,K\left.\right)$ in a spherical can, separated by a vacuum space from a surrounding shield which is maintained at the temperature of liquid nitrogen $\left(\right.77\,K\left.\right)$ . If the can is $0.1\,m$ in radius and is blacked on the outside so that it acts as a black body, how much helium boils away per hour?
(Latent heat of vapourization is $21\,kJ\,kg^{- 1}$ )

• A. $43\,gh^{- 1}$
• B. $43\,kgh^{- 1}$
• C. $4.3\,gh^{- 1}$
• D. $43\times 10^{- 3}gh^{- 1}$

Answer 1

Answer: (A)

Heat absorbed per second by liquid helium $\left(T_{0}^{4}-T^{4}\right)$
Heat required to boil liquid helium away $= \left(\frac{\text{dm}}{\text{dt}}\right) \text{L}$
$\left(\frac{ dm }{ dt }\right)=\frac{\sigma A \left( T _{0}^{4}- T ^{4}\right)}{ L }$
$=\frac{6.67 \times 10^{-8} \times 4 \times 3.14 \times(0.1)^{2}\left(77^{4}-4.2^{4}\right)}{21 \times 10^{3}}$
$= \text{1.19} \times 1 0^{- 5} kg/s$
$= \text{4.3} \times 1 0^{- 2} kg/h = 4 3 g/h$