Question

Liquid helium is stored at its boiling point (4.2 K) in a spherical can, separated by a vacuum space from a surrounding shield which is maintained at the temperature of liquid nitrogen (77 K). If the can is 0.1m 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)

• A. 43 g/h
• B. 43 kg/h
• C. 4.3 g/h
• D. 43 x 10^-3 g/h

Answer 1

Answer: (A)

Heat absorbed per second by liquid helium $\alpha A=\left(T_0^4-T^4\right)$
Heat required to boil liquid helium away $=\left(\frac{dm}{dt}\right)L$
$\begin{array}{rl}<br/><br/>(\frac{dm}{dt}& )=\frac{\sigma A\left(T_0^4-T^4\right)}{L}\\ <br/><br/>& =\frac{6.67\times 10^{-8}\times 4\times 3.14\times (0.1{)}^2\left(77^4-4.2^4\right)}{21\times 10^3}\\ <br/><br/>& =1.19\times 10^{-5}kg/s\\ <br/><br/>& =4.3\times 10^{-2}kg/h=43g/h<br/><br/>\end{array}$