Question

A physicist uses a cylindrical metal can 0.230 m high and 0.0850 m in diameter to store liquid helium at 4.22 K; at that temperature the heat of vaporization of helium is 2.09×104J/kg. Completely surrounding the metal can are walls maintained at the temperature of liquid nitrogen, 77.3 K, with vacuum between the can and the surrounding walls.

Part A How much helium is lost per hour? The emissivity of the metal can is 0.200. The only heat transfer between the metal can and the surrounding walls is by radiation.

Answer 1

A)

h = height of the cylinder = 0.230 m

d = diameter of cylinder = 0.0850 m

r = radius of cylinder = d/2 = 0.085/2 = 0.0425 m

Area of cylinder is given as

A = 2r(r + h) = 2 (3.14) (0.0425) (0.0425 + 0.230) = 0.07273 m²

T_h = Temperature of helium = 4.22 K

T_s = Temperature of surrounding = 77.3 K

= emissivity = 0.2

Heat of radiation is given as

Q_r = A (T_h⁴ - T_s⁴)

Q_r = (5.67 x 10^-8)(0.2) (0.07273) (77.3⁴ - 4.22⁴)

Q_r = 0.02945

Using conservation of heat

heat of vapourization = heat of radiation

mL = Q_r t

Rate of loss of helium is given as

m/t = Q_r /L

m/t = 0.02945/(2.09 x 10⁴)

m/t = 1.41 x 10^-6 kg/s

m/t = 1.41 x 10^-6 x 3600 kg/h

m/t = 5.076 x 10^-3 kg/h

m/t = 5.076 g/h