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 0.1 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 )

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 0.1 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× 10- 3gh- 1

Tardigrade Admin · Accepted Answer

Heat absorbed per second by liquid helium

(T_{0}^{4} - T^{4})

Heat required to boil liquid helium away

= (\frac{dm}{dt}) L

(\frac{d m}{d t}) = \frac{σ A ( T _{0}^{4} - T ^{4} )}{L}

= \frac{6.67 \times 1 0 ^{- 8} \times 4 \times 3.14 \times ( 0.1 ) ^{2} ( 7 7 ^{4} - 4. 2 ^{4} )}{21 \times 1 0 ^{3}}

= 1.19 \times 1 0^{- 5} k g / s

= 4.3 \times 1 0^{- 2} k g / h = 43 g / h

$43 g h^{- 1}$

$43 k g h^{- 1}$

$4.3 g h^{- 1}$

$43 \times 1 0^{- 3} g h^{- 1}$

43gh−1

43kgh−1

4.3gh−1

43×10−3gh−1

Heat absorbed per second by liquid helium (T04​−T4) Heat required to boil liquid helium away =(dtdm​)L (dtdm​)=LσA(T04​−T4)​ =21×1036.67×10−8×4×3.14×(0.1)2(774−4.24)​ =1.19×10−5kg/s =4.3×10−2kg/h=43g/h

$43 g h^{- 1}$

$43 k g h^{- 1}$

$4.3 g h^{- 1}$

$43 \times 1 0^{- 3} g h^{- 1}$