A cylindrical dewar has a height of L=25 cm and a volume of V = 1 liter, but the

Question

A cylindrical dewar has a height of L=25 cm and a volume of V = 1 liter, but the manufacturer goofed and forgot to apply silver films to the pyrex-glass vacuum walls. (Pyrex has an emissivity e = 0.9.) The dewar is initially filled completely with liquid nitrogen at its boiling point = 77 K. at 1 atm. (The density of liquid nitrogen is 800 kg/m3 and its latent heat of vaporization at 1 at. Lv = 1.99 x 105 J/kg). Consider a day when the outside temperature = 20 o C. a) What is the rate that energy is transferred by radiation between the inner and outer vacuum walls? [For the area of the container, it is OK to just consider the lateral area, A = 2?rL.] b) How long will it take for all the nitrogen to evaporate if radiation is the only source of heat and the inner vacuum walls remain at 77K as the nitrogen evaporates?

Explanation / Answer

Volume of the cylinder = V = pi R^2 L

Q1 = Heat of vaporization = mLv

Mass of liquid nitrogen = m = density x volume = 800 x 0.001 = 0.8 kg

Q1 = (0.8)(1.99 e5) = 1.592 e5 J

According to Stefan-Boltzmann law, rate of energy transfer P= eA (T^4-Tc^4)

P= (0.9)(5.67 x10^-8)(2 pi *0.0357*0.25)( 293^4 -77^4) = 20.98 J/s

b) Since the Nitrogen is at the boiling point , the heat energy transferred is just

Q1= mLv and none goes to the flask.

= 1.592 x 10^5 J

Time = energy / power = (1.592 x10^5)/(20.98)

= 7588 seconds

= 2.1 hours

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