An aluminum rod 0.500 m in length and with a cross-sectional area of 2.40 cm2 is

Question

An aluminum rod 0.500 m in length and with a cross-sectional area of 2.40 cm2 is inserted into a thermally insulated vessel containing liquid helium at 4.20 K. The rod is initially at 340 K. (Aluminum has thermal conductivity of 3 100 W/m . K at 4.20 K; ignore its temperature variation. Aluminum has a specific heat of 0.215 and density of 2.70 g/cm3. The density of liquid helium is 0.125 g/cm3.) (a) 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. liters (b) If the circular surface of the upper end of the rod is maintained at 340 K, what is the approximate boil-off rate of liquid helium after the lower half has reached 4.20 K? liters/s A 600-kg meteoroid happens to be composed of aluminum. When it is far from the Earth, its temperature is and it moves at 13.5 km/s relative to the planet. As it crashes into the Earth, assume the internal energy transformed from the mechanical energy of the meteoroid-Earth system is shared equally between the meteoroid and the Earth and all the material of the meteoroid rises momentarily to the same final temperature. Find this temperature. (Assume the specific heat of liquid and of gaseous aluminum is, the melting point is, the boiling point is, the latent heat of fusion is 3.97 x 105 J/kg, and the latent heat of vaporization is 1.14 x 107 J/kg.)

Explanation / Answer

4.)

a.)(pVLv)He = (pVcdT)Al

So V He = (pVcdT)Al/(pLv)He = (2.7)(62.5)(900)(335.8)/(0.125)(2.09x10^4)= 19521.4 cm^3 = 19.5 Lt

b.)dQ/dt=P=KAdT/L

dQ/dt=(dm/dt)Lv

So dm/dt=KAdT/L.Lv

So dm/dt=(3100/100)(2.4)(335.8)/(25)(20.9)= 47.815 g/s

dV/dt=47.815/0.125 = 382.5 cm^3/s = 0.382 Lt/s

5.)

The K of the meteorite = 1/2*m*v^2 = 1/2*600kg*(13500m/s)^2 = 5.4675 x 10^10 J

The final temp will be calculated in 5 steps possibly

1) Raise temp to the melting point
using Q = m*c*?T = 600kg * 910J/kg-oC*(660 + 20) = 3.71x10^8 J

2) Melt the Al using Q = m*Lf = 600kg*3.97x10^5J/kg = 2.38x10^8J

3) raise temp to boiling point using

Q = m*c*?T = 600*1170*(2450 -660) = 1.256x10^9 J

4) Vaporize the Al using Q = m*Lv = 600*1.14x10^7J/kg = 6.84x10^9J

so far the total energy used is the sum of the above = 8.7x10^9J

so there are 5.4675 x 10^10 - 8.7x10^9 = 4.597x10^10J remaining to raise the temp to the final value

So we have 4.597x10^10 = m*c*(Tf - 2450) = 600*1170*(Tf - 2450)

So Tf= 67934.33 C

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