An aluminum rod 0.500 m in length and with a cross-sectional area of 2.30 cm^2 i

Question

An aluminum rod 0.500 m in length and with a cross-sectional area of 2.30 cm^2 is inserted into a thermally insulated vessel containing liquid helium at 4.20 K. The rod Is initially at 365 K. (Aluminum has thermal conductivity of 3 100 W/m middot K at 4.20 K; ignore its temperature variation. Aluminum has a specific heat of 0.215 cal/g middot degree C and density of 2.70 g/cm^3. The density of liquid helium is 0.125 g/cm^3.) If one-half of the rod is inserted into the helium, how many liters of helium boil off by the time the inserted half cools to 4.20 K? Assume the upper half does not yet cool. If the circular surface of the upper end of the rod is maintained at 365 K, what is the approximate boil-off rate of liquid helium after the lower half has reached 4.20 K? Demonstrations with liquid nitrogen give us some indication of the phenomenon described. Since the rod is much hotter than the liquid helium and it is of significant size, a substantial volume (maybe as much as a liter) of helium will boil off before thermal equilibrium is reached. Likewise, since aluminum conducts rather well, a significant amount of helium will continue to boil off as long as the upper end of the rod is maintained at the higher temperature. Until thermal equilibrium is reached, the excess internal energy of the rod will go into vaporizing liquid helium, which is already at its boiling point, so there is no change in the temperature of the helium.

Explanation / Answer

Q1& 2. Heat tranfered from aluminium rod =

Rate = dQ/dt = k * A * dT/dx

Since heat transfered is absorbed by helium. Latent heat of vapourization of helium.So L * dm /dt = k * A * dT/dx

So dm/dt = k * A * (365 - 4.2) / L* l/2 wherel isthe length of aluminium rod.

Now After calculate rate of helium boiling.

Wealso know that for aluminium rate of heat tranfer is m* s * dT where s is the specific heat.

So m * s * dT / ( 325 - T) = k * A * dt / (l/2)

So temperature falls exponentially wiyh time. After solving theeqn, and substitutinf final temperture as 4 . to get the time required to cool. This will give how much time helium has been boiling and how much heat has been trasnferred. Q net = mL. required for total helium boiled.

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