A 850 -kg meteoroid happens to be composed of aluminum. When it is far from the

Question

A 850-kg meteoroid happens to be composed of aluminum. When it is far from the Earth, its temperature is -18.5

A 850-kg meteoroid happens to be composed of aluminum. When it is far from the Earth, its temperature is -18.5 degree C and it moves at 14.5 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 1 170 J/kg · degree C, the melting point is 660 degree C, the boiling point is 2 450 degree C, the latent heat of fusion is 3.97 105 J/kg, and the latent heat of vaporization is 1.14 107 J/kg.) C

Explanation / Answer

given that mass of meteroid m = 850 kg temperature t = -18.50 c speed 14.5 km/s specific heat = 1170 J/kg.c

the lost of mechanical energy is =( 1/2) mvi2 + GME*m/RE = ( 1/2) 850 ( 1.45 x 10 ^4 m/s)^2 + 6.67 x 10^-11 ( 5.98 x 10^24 ) ( 850 )/ 6.37 x 10^6 = 8.93 x 10^10 J +5.32 x 10^10 J = 14.25 x 10^10 J one half becomes extra internal energy in hte aluminimum ?Eint = 7.125 x 10^10 J to raise temperature to the mealting point requires energy mc?T = 850 kg ( 900 J/kg.c) (660 -(- 18.5) = 5.19 x 10^8 J to melt mL =850 x ( 3.97 x 10 ^5 kg) = 3.374 x 10^8 J to raise it to be the boiling point mc?T = 850 ( 1170 ) ( 2450 - 660 ) = 1.78 x 10^9 J to boil it mL =850 kg ( 1.14 x 10^7 J/kg) = 9.7 x 10 ^9 J then 7.125 x 10^10 J = 12.34 x 10^9 J + 850 ( 1170) ( T(f) - 2450 ) J/c T(f) = 6.168 x 10^4 C0 given that mass of meteroid m = 850 kg temperature t = -18.50 c speed 14.5 km/s specific heat = 1170 J/kg.c

the lost of mechanical energy is =( 1/2) mvi2 + GME*m/RE = ( 1/2) 850 ( 1.45 x 10 ^4 m/s)^2 + 6.67 x 10^-11 ( 5.98 x 10^24 ) ( 850 )/ 6.37 x 10^6 = 8.93 x 10^10 J +5.32 x 10^10 J = 14.25 x 10^10 J one half becomes extra internal energy in hte aluminimum ?Eint = 7.125 x 10^10 J to raise temperature to the mealting point requires energy mc?T = 850 kg ( 900 J/kg.c) (660 -(- 18.5) = 5.19 x 10^8 J to melt mL =850 x ( 3.97 x 10 ^5 kg) = 3.374 x 10^8 J to raise it to be the boiling point mc?T = 850 ( 1170 ) ( 2450 - 660 ) = 1.78 x 10^9 J to boil it mL =850 kg ( 1.14 x 10^7 J/kg) = 9.7 x 10 ^9 J then 7.125 x 10^10 J = 12.34 x 10^9 J + 850 ( 1170) ( T(f) - 2450 ) J/c T(f) = 6.168 x 10^4 C0 = ( 1/2) 850 ( 1.45 x 10 ^4 m/s)^2 + 6.67 x 10^-11 ( 5.98 x 10^24 ) ( 850 )/ 6.37 x 10^6 = 8.93 x 10^10 J +5.32 x 10^10 J = 14.25 x 10^10 J one half becomes extra internal energy in hte aluminimum ?Eint = 7.125 x 10^10 J to raise temperature to the mealting point requires energy mc?T = 850 kg ( 900 J/kg.c) (660 -(- 18.5) = 5.19 x 10^8 J to melt mL =850 x ( 3.97 x 10 ^5 kg) = 3.374 x 10^8 J to raise it to be the boiling point mc?T = 850 ( 1170 ) ( 2450 - 660 ) = 1.78 x 10^9 J to boil it mL =850 kg ( 1.14 x 10^7 J/kg) = 9.7 x 10 ^9 J then 7.125 x 10^10 J = 12.34 x 10^9 J + 850 ( 1170) ( T(f) - 2450 ) J/c T(f) = 6.168 x 10^4 C0

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