Question

A cylindrical metal can, 0.1 m high and 0.05 m in diameter, contains liquid helium at its normal boiling point of -452.074 degrees Fahrenheit. At this temperature Helium's heat of vaporization is 20.4 kJ/kg. The walls of the helium container are 1.2 cm thick and have a thermal conductivity of 13.889 W/(m K). The helium container is surrounded by liquid nitrogen at a temperature of -327.64 degrees Fahrenheit.

A. What is the conductive surface area of the metal cylinder?

B. What is the temperature of the liquid helium in Kelvin?

C. What is the temperature of the walls in Kelvin?

D. What is the rate of heat flow into the Helium due to the conduction?

E. How much Helium is lost per hour due to conduction?

Answer 1

part.(A)

Conducting surface area of metal cylinder = 0.1 0.05 m² = 0.0157 m²

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part (B) Temperature of liquid Helium in K :- ( -452.074 - 32 ) (5/9) + 273 = 4.07 K

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part(C) Temperature of liquid Nitrogen in K :- ( -327.64 -32 ) (5/9) + 273 = 73.2K

walls are at temperature of liquid helium and liquid nitrogen , i.e., 4.07 K and 73.2 K

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Rate of heat flow due to conduction = k A (dT/dl)

where k = 13.889 W/mK , thermal conductivity of material of can

A is area of the conducting surface and (dT/dl) is temperature gradient across the wall of cotainer

Rate of heat flow due to conduction = 13.889 0.0157 ( 73.2 - 4.07 ) / 0.012 = 1256 W

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Thermal energy transferred in an hour = 1256 3600 = 4522 kJ

Helium lost per hour = 4522 / 20.4 222 kg

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